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Elements of 
Abstract Harmonic 
Analysis 



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Elements of 
Abstract Harmonic 
Analysis 


By George Bachman 

POLYTECHNIC INSTITUTE OF BROOKLYN 
DEPARTMENT OF MATHEMATICS 
BROOKLYN, NEW YORK 


with the assistance of 
Lawrence Narici 

POLYTECHNIC INSTITUTE OF BROOKLYN 
DEPARTMENT OF MATHEMATICS 
BROOKLYN, NEW YORK 



ACADEMIC PRESS • New York and London 



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Preface 


Abstract Harmonic Analysis is an active branch of modern 
analysis which is increasing in importance as a standard 
course for the beginning graduate student. Concepts like 
Banach algebras, Haar measure, locally compact Abelian 
groups, etc., appear in many current research papers. This 
book is intended to enable the student to approach the 
original literature more quickly by informing him of these 
concepts and the basic theorems involving them. 

In order to give a reasonably complete and self-contained 
introduction to the subject, most of the proofs have been 
presented in great detail thereby making the development 
understandable to a very wide audience. Exercises have been 
supplied at the end of each chapter. Some of these are meant 
to extend the theory slightly while others should serve to 
test the reader’s understanding of the material presented. 

The first chapter and part of the second give a brief 
review of classical Fourier analysis and present concepts 
which will subsequently be generalized to a more abstract 
framework. The presentation of this material is not meant 
to be detailed but is given mainly to motivate the generaliza- 
tions obtained later in the book. The next five chapters pre- 
sent an introduction to commutative Banach algebras, gen- 
eral topological spaces, and topological groups. We hope that 
Chapters 2-6 might serve as an adequate introduction for 
those students primarily interested in the theory of com- 
mutative Banach algebras as well as serving as needed pre- 
requisite material for the abstract harmonic analysis. The 
remaining chapters contain some of the -measure theoretic 
background, including the Haar integral, and an extension of 
the concepts of the first two chapters to Fourier analysis on 
locally compact topological abelian groups. 

In an attempt to make the book as self-contained and as 
introductory as possible, it was felt advisable to start from 
scratch with many concepts — in particular with general 



Preface 


topological spaces However within the space limitations 
it was not possible to do this with certain other background 
material— notably some measure theory and a few facts from 
functional analysis Nevertheless the material needed from 
these areas has all been listed in various appendices to the 
chapters to which they are most relevant 

There are now a number of more advanced books on 
abstract harmonic analysis which go deeper into the subject 
We cite in particular the references to Rudin Loomis and 
the recent book by Hewitt and Ross Our treatment of the 
latter part of Chapter 12 follows to some extent the initial 
chapter of the book by Rudin which would be an excellent 
continuation for the reader who wishes to pursue these 
matters further 

The present book is based on a one semester course m 
abstract harmonic analysis given at the Polytechnic In 
stitute of Brooklyn during the summer of 1963 for which 
lecture notes were written by Lawrence Narici A few re 
visions and expansions have been made 

I would like to express my sincere gratitude to Mr Nanci 
for bis effort in writing the notes improving many of the 
proofs and for editing the entire manuscript I would like 
finally to express also my appreciation to Melvin Maron for 
his help m the preparation of the manuscript 


George Bachman 



Contents 

PREFACE V 

SYMBOLS USED IN TEXT X 

Chapter 1 The Fourier Transform on the Real Line for 

Functions in Li 1 

Introduction 1 

Notation 1 

The Fourier Transform 2 

Recovery 4 

Relation between the Norms of the Fourier Transform and the 

Function 10 

Appendix to Chapter 1 15 

Exercises 17 

References IS 

Chapter 2 The Fourier Transform on the Real Line for 

Functions in l 2 19 

Inversion in hi 21 

Normed and Banach Algebras 25 

Analytic Properties of Functions from C into Banach Algebras 29 

Exercise 33 

References 33 

Chapter 3 Regular Points and Spectrum 34 

Compactness of the Spectrum 38 

Introduction to the Gel’fand Theory of Commutative Banach 

Algebras 48 

The Quotient Algebra 50 

Exercises 53 

References 54 

Chapter 4 More on the Gel'fand Theory and an Intro- 
duction to Point Set Topology 55 

Topology 60 

A Topological Space 61 

vii 



Contents 


Examples of Topological Spaces 61 

Further Topological Notions 62 

The Neighborhood Approach 66 

Exercises 71 

References 72 

Chapter 5 Further Topological Notion* 73 

Bases, Fundamental Systems of Neighborhoods, and Subbases 73 
The Relative Topology and Product Spaces 78 

Separation Axioms and Compactness 79 

The Tychonoff Theorem and Locally Compact Spaces 84 

A Neighborhood Topology for the Set of Maximal Ideals over a 
Banach Algebra 87 

Exercises 89 

References 90 

Chapter 6 Compactness of the Space of Maximal Ideals 
over a Banach Algebra; an Introduction to 
Topological Groups and Star Algebras 91 

Star Algebras 97 

Topological Groups 98 

Exercises 106 

References 106 

Chapter 7 The Quotient Group of a Topological Crovp 

and Some Further Topological Notions 107 

Locally Compact Topological Groups 107 

Subgroups and Quotient Groups 109 

Directed Sets and Generalized Sequences 116 

Further Topological Notions 117 

Exercises 123 

References 124 

Chapter 8 Right Haar Measures and the Haar Covering 

Function 125 

Notation and Some Measure Theoretic Results 125 

The Haar Covering Function 129 

Summary of Theorems in Chapter 8 147 

Exercises 149 

References 149 


Contents ix 

Chapter 9 The Existence of a Right invariant Haar 
Integral over any Locally Compact Topolog- 
ical Group 150 

The Daniell Extension Approach 158 

A Measure Theoretic Approach 160 

Appendix to Chapter 9 163 

Exercises 164 

References 165 

Chapter 10 The Daniell Extension from a Topological 
Point of View, Some General Results from 
Measure Theory, and Group Algebras 166 

Extending the Integral 166 

Uniqueness of the Integral 169 

Examples of Haar Measures 172 

Product Measures 176 

Exercises 186 

References 187 

Chapter 1 1 Characters and the Dual Group of a Locally 

Compact, Abelian, Topological Group 188 

Characters and the Dual Group 192 

Examples of Characters 200 

Exercises 206 

References 206 

Chapter 12 Generalization of the Fourier Transform to 

Li(G) and L 2 (G) 20/ 

The Fourier Transform on Li(G) 207 

Complex Measures 211 

The Fourier-Stieltjes Transform 219 

Positive Definite Functions 220 

The Fourier Transform on Lj(G) 235 

Exercises 243 

Appendix to Chapter 12 244 

References 250 

BIBLIOGRAPHY 251 


INDEX 


253 



Symbols Used in Text 


\x i pi 
/ X —*Y 
x-*y 


CE 

0 

c 


The set of all x with property p 
The function / mapping the set X into the set Y 
where x £ X is mapped into y 6 Y Further, if 
E C X,f(E) will denote the image set of E 
under /, ie, /(£) =* j/{x) J x € E) If 
M C will denote those x € X such 

that /(x) € 21/ The notation / |b, E C X, will 
denote the restriction of / to E 
The complement of the set E 
The null set 
The complex numbers 


In the list below the number immediately following the 
symbol will denote the page on which the symbol is defined 
The symbols are bsted according to the order m which they 
appear in the text 


R, 1 
Ep, 1 

11/ Ik 2, J)/ IJ., 180, 
11/ II., 196 
/ ' g, 6, 180 
/, 2 

L(X, X), 26 
Li(Z), 28 

|M ||, A€L(X,X), 26 

C£a, bj, 26 

W, 26 

Z, 28 

<r(x), 37 

Rx, 41 

r.(x), 42 

M, 52 

x(M), M €. M, 51 
F(&), 57 


6, 61 
E, 62 
E°, 71 
V{x), 66 

V{M,x h x t , •••,*., «), 87 
C*, 99 
R*, 99 

ViV t - |xy|z€ V,,y€ V,}, 
100 

[/-* = ix~'\x€ V), 100 
G/H, 110 
lim /(s), 116 
s 

X*, 118 
C, 125 
ft 126 
C«(G), 129 



C 0 + (G), 129 
k E , 129 
(/:*), 130 

I, if), 136 
I if), 157 
/**, 161 
E z , E y , 177 
f x , U 177, 188 
U{G), 180 


Symbols Used in Text 


xt 


f*(x), 188 

R(G), 193 
x, 193 
fix), 193 
N( X o,E, e), 204 
L 2 (G), 220 
B(X), B(Y), 212 
fi * v, 214 
Nix 0 ,E,e), 222 



CHAPTER 7 


The Fourier Transform on the 
Real Line for Functions in Li 


Introduction 

This section will be devoted to some Fourier analysis on 
the real line. The notion of convolution will be introduced 
and some relationships between functions and their trans- 
forms will be derived. Some ideas from real variables (e.g. 
passage to the limit under the Lebesgue integral and inter- 
changing the order of integration in iterated integrals) will 
be heavily relied upon and a summary of certain theorems 
that will be used extensively in this section are included in 
the appendix at the end of this chapter. 


Notation 

Throughout, R will denote the real axis, ( — «> , a> ) , and / 
will denote a measurable function on R (f may he real or 
complex valued) . We will denote by L p the set of all meas- 
urable functions on R with the property! 

| f(x) \ F dx < co , 1 < p < CO. 

Also, since we will so frequently be integrating from minus 
infinity to plus infinity, we will denote 

f by simply f 

^ — co J 

where all integrations are taken in the Lebesgue sense. 



t One often says that / is pth power summable. 



Element* of Abstract Harmonic Analysis 


Noting without proof that the space L p is a linear space 
we can now define the norm of f with respect to p, / j) p , where 
/ 6 L p as follows 

11/11, = (/ 1 m I ’**)'" 

It is simple to verify that the following assertions are valid 

1 I! /II* ^ 0 and |( /(ip = 0 if and only if f ~ Ot 

2 II VII, “ Ul ll/ll, Where l is a real or complex 
number Note that this immediately implies (1/ |j p = 
11 -/ 11 * 

3 11/ + 9 lie < ll/ll, + II 9 ll» To verify this all one 
needs is Minkowski’s inequality for integrals which is 
listed m the appendix to this chapter (p 15) 

Thus it is seen that L„ is a normed linear space One can 
go one step further however although it will not be proven 
here it turns out that L, is actually a complete normed linear 
space or Banach space or that any Cauchy sequence (in the 
norm with respect to p) of functions in L p comerges (in the 
norm with respect to p) to a function that is pth power 
summable 


The Fourier Tronsform 

In this section we consider the Fourier transform of a 
function / in Li and note certain properties of the Fourier 
transform Let / 6 Lj and consider as the Founer transform 

off 

m = /,■"/(() a 


t We mil say that two functions / and q are equivalent, denoted by 
/ ~ 0, if / = g almost everywhere Thus I P is actually the set of all 
equivalence classes of functions tinder this equivalence relation If we 
’utA, tills, to 1) -wttoWi TtprtiiTA. ■& paeudw.cm because V. 
would be possible for J)/)|> to equal zero even though/ was not zero 



1. Fourier Transform for Functions in Li 


3 


We first note that f(x) exists for, since / 6 L h 
I /(*) I < / \m\dt < oo. 

Further, since 

/ 1/(0 I* = II / IK 

we can say 

\f(x) I < 11/ 111 

for any x ; or that the 1-norm of / is an upper bound for the 
Fourier transform of /. Since it is an upper bound it is cer- 
tainly greater than or equal to the least upper bound or 

sup | J{x) | < || /||i. 

xzR 

It will now be shown that the Fourier transform of / is a 
continuous function of x. Consider the difference 

fix + h„) - f{x) = J e' xi (e ih "‘ - l)/(<) dt 
where h„ € R: 

\f(x + h„) - f(x) | < J | expiihj) - 1 | | fit) | dt. 

Since this is true for any h„ it is true in the limit as h n ap- 
proaches zero, or 

lim (/(s + h„) — fix) | < lim f | e ihnt — 1 | | /(<) | dt. 

kn~+f) hn~*0 

Our wish now is to take the limit operation under the inte- 
gral sign, and Lebesgue's dominated convergence theorem 
will allow this manipulation (see appendix to Chapter 1) for 
certainly the integrand in the last expression is dominated 



dementi of Abstract Harmonic Analysis 


by the summable function 2 |/(f) | Taking the limit inside 
yields the desired result, namely, that 

lim /(* + K) - /(x) 

or that the Fourier transform / of a function / m I«i is a con- 
tinuous function 

The following fact can also be demonstrated about /(x) 
and is usually referred to as the Riemann-Lebesgue lemma 
lim /( jc) = 0 (1) 

We note in passing that there are continuous functions, 
F(i), satisfying (1) but such that no /(<) can be found that 
satisfies 

F(i) - j «“'/(<) dt 

This bnngs us to our next problem knowing /(x), how can 
the function that it came from,/(t), be found again? 

Recovery 

In elementary treatments one often sees the following 
inversion formula 

/(<) “ ~ /' dz 

It will now be demonstrated, by a counterexample, that the 
above formula is not true in general 

Example Consider the function 

(e~‘, t > 0 

m = i 

ip, t < 0 

/(x) - f '%«-»■<« - — 

J o XX — 1 



1 . Fourier Transform for Functions in L j 


5 


It is now clearly impossible to recover /(f) using the formula 
for 

/ | | lx - /-^=_ . 

Since the last integrand behaves like 1/a:, for large x the inte- 
gral will become infinite as log x and, therefore, we cannot 
recover /(f) using that formula. (Note that a Lebesgue 
integral must be absolutely convergent in order to converge.) 

Before proceeding further, two results from real variables 
will be needed: 

Definition, t is said to be a Lebesgue point of the function f if 
1 f t+h 

iim i / I /(«) ~ f(x) 1 dx = 0. 

A-0 “ J t 

Theorem. If / € L lt then almost all of its points are Lebesgue 
points. 

Theorem. Every point of continuity of a function is a 
Lebesgue point. 

The following two theorems on inversion are lengthy and 
somewhat intricate and the reader is referred to Goldberg 
[ ’2] for the proofs. 

Theorem 1 . Let /, / 6 Li and suppose / is continuous at t, 
then 

/(f) = — f er** l f(x) dx. 

Theorem 2. Let f € Li and let f be a Lebesgue point for the 
function, /, then 

/(f) = lim — f (l - — V fa '/(aO dx. 

(Note that this limiting process is analogous to ( C , 1) sum- 
mability criterion for infinite series.) 



Elements of Abstract Hormonic Anolysu 


Corollary 1. Suppose / 6 Za and f(x) — 0, for all x, then 
f(t) = 0,ae 

Corollary 2. Suppose/i,/i £ L\ If /i = then/i(t) = /i(t) 
ae 

This result follows immediately from Corollary 1, for we 
have 

/\ 

u-u-h -/. -o 

therefore 

/ir/i B 0. a e 

Convolution Let f g € Li and consider the function 

fc(x) - f}(x - 1)9(1) <fl = (/-sKi) 

and called the conuolufton o/ / tattfc p It is now contended 
that k(z) exists for almost all x and that k(z) is summable 

Proof It is easily shown that 

ff{x - t)dx = Jf{ x) dx (2) 

by simply making a change of variable Now consider 
fdtf |/(x - 1)9(1) I dx - / | j(l) | dlf |/<x - I) | dx 
-Il9ll.il/ll. < ” 

by usmg (2) Now, by the Tonelli-IIobson theorem (see 
appendix to Chapter 1), it follows that 

jjf(x - t)g(t) dtdx 

is absolutely convergent By the Fubim theorem it follows 
that h(x) exists a e and is integrable 
It will now be shown that the operation of convolution is a 
commutative one 

Theorem / » g = g */for /, g £ Z* 



1. Fourier Transform for Functions in Li 


7 


Proof. 


if *g) (x) = Jf(x ~ t)g(t) dt. 

Let u = x — t. Then 

(f * (j) ( x ) = / f(u)g(x - u)(-du ) = (<?*/) (s). 

~ CO 

It also follows that the operation of convolution is associa- 
tive or 

f*(g*h) = U*g) *h 

where /, g, h 6 Lj. 

The proof of this result, although straightforward, is 
rather messy and will be omitted. 

Theorem. Let f, g (• la. Then 

\\f*g\\i < 11/ Hill g Ili- 

Proof. 

II/* S' Hi = Jdx\ Jf(x - t)g(t) dt | 

< fdxj |/(x - t)g(t) | dt. (3) 

We now note that 

Jdtj | f(x - t)g(t ) | dx = J | g{t) | dtj | f(x - t) | dx 
= lUllill/lli < 00 ( 4 ) 

or that (3) converges absolutely. By the Tonelli-Hobson 
theorem,! 

Jdxj | f{x — t )g{t) | dt 

converges absolutely and, by the same theorem must also 


t See appendix to Chapter 1. 



Element] of Abstract Harmonic Analysis 


converge absolutely to the value 

jdtf |/(l - t)j(i) \it - l!/|!i(UH, 

which yields the desired result 
We can summarize the above results by noting that la 
with respect to the operations of addition and convolution 
forms a Banach algebra (see Chapter 2, Example 6) 

The next theorem is one of the mam reasons for interest 
in convolution 

Theorem. Let / g € Lj, then 

/\ , 

/ • » ~ a 

Proof 

/\ 

</ • 5) (*) 


Since 

fdsf |e“'/(t - 


=* dl 

«=> je al dtjf(t - s)g{s) ds (5) 

s)g(s)\dt=* jdej \f(t ~ s)g(s) \ dt 


= / \9{*)\dsj 

« II 9 Hill /Hi < • 

then, by the same reasonuig as m the previous theorem (use 
of the TonelU-Hobson theorem), the order of integration 
may be interchanged in (5) Wnte 
e u ‘ = 



1. Fourier Transform for Functions in Li 


9 


substitute this in (5), and interchange the order of integra- 
tion. Now 

(/* 9 ) 0*0 = / ff (•)«** ds Jf(t - s)e' z "-‘ ) dt= gf = fg. QED 

We noted that Li(+, *) was a Banach algebra. We will now 
show that it is not an algebra with identity for suppose 
there were an element, e € Lj, such that 

f *e — J for every / 6 L 2 . 

Then certainly 

e * e = e. 

By the previous theorem, then 

/\ 

e * e = — e. 

Therefore t — 0, 1 are the only possibilities. By continuity 
of 8, e must equal either zero or one; it cannot jump! But, 
since we require in addition 

lim e(x) = 0, 

X-*CO 

we must choose 

i(x) = 0. 

This implies that e(f) = 0, a.e. and the only way that this 
could be an identity would be if f(t) = 0, a.e., for all / 6 la, 
which is ridiculous. 

Even though there can be no identity there is what is 
called an approximate identity, there exists a sequence of 
functions, {e„} in Li such that lim„^„ || e n *f — / ||i = O.f 


t See Goldberg [2], 



?0 


Elements of Abstract Harmonic Analysis 


Relation between the Norms of the Fourier Transform 
and the Function 

Before proceeding to the main result we need the following 
Lemmo 1 Let a, b 6 B, b > 0 Then 

/«"'exp(-W’) it - (0 ' exp (-f) 

The proof will only be outlined 

Consider J exp (— 2 *) dz where z is a complex variable 
and F ts the contour shown below 



Splitting up the integral into 

and taking the limit as k — * «> yields the desired result We 
must also establish the following limit before proceeding to 
the theorem 

Lemma 2 Let/ € Li n L t , then 

lim j exp (—**/«) )/(*) ]*di = 2ir ]) / ))| 



T . Fourier Transform for Functions in ii 


11 


Proof. Certainly 

I /(*) I 2 = f(x)](x) 

= J f(t)e ixl dtj f(s)er ix ‘ ds. 

Multiplying both sides by exp ( — x 2 /n) , where n is an integer, 
and integrating with respect to x, we have 

J exp(—x 2 /n) ) f(x) | 2 da: 

= J exp (—x 2 /n)dxjf(t)e' x, dtjf(s)e~ ix ’ds. 

By the exact same reasoning used before we can interchange 
the order of integration to get 

y"exp( — x 2 /n) | f(x) | 2 dx 

= ff(s) ds jf(t) dt Je ix(t ~’ } exp(— x 2 /n) dx. 

Using the result of Lemma 1 with a — t — s, b = 1/n, x = t 
we can evaluate the last integral on the right as 

y^ e ivr(i-j) exp(— x 2 /n) dx — (mi) 112 exp( — n(t — s) 2 /4) 

or 

J exp (— x 2 /n) [ f(x) \ 2 dx 

= (xn)» 2 J/( S ) dsjf(t) exp (-n(i - s) 2 /4) dt. (*) 



12 


Element* of Absfroct Harmonic Analysis 


Interchanging the order of integration [see Eq («) below 
for the justification]] we get 

j exp (-!*/«) l/(x) \ a dx 

«= (t n)* n j exp (— n£*/4) dtjf(t + s)/(s) ds 

after first replacing / by t + s in (^) 

Let us denote by g(t) 

9(0 - //« + *)/(<) d* 

or 

j exp(— z J /n) !/(*) |* dr = (s exp (— nt 1 / 4 ) * 
Replace / by 2n~ w / This gives for the preceding integral 
2r'” J g(2n~ ll H) exp( -t 1 ) dt 

We now claim that g(t) is continuous at the origin Hence 
we must show 

lim | g(t ) - g( 0 ) | =0 

I 9(0 - 9(0) |> - | //(»)(/(< + «) -/(«)) *1* 

< / |/<5) |’<b- / |/(l + *) -/W I'*. 

and since 

hm j j f(t + s) - /(«) | l ds = 0 



1. Fourier Transform for Functions in L i 


13 


(see exercise 4) we therefore have 

lim | g(l) - g(0) | = 0. 

t -+ 0 

Thus g(t) is continuous at the origin. Now 

I g(t) I = J ff(t + s)f(s) ds 

< J \f(t + s)f(s) | ds 

< (/ !/ ( t + s) |* *)”*(/ l/(s) \ 2 dsJ 2 
by the Cauchy-Schwarz inequality. Thus 

MO I < 11/11*11/11* = 11/115 <•) 

for any t. Now we can say for any n and, in particular, in 
the limit as n — * oo , 

lim J exp (— x 2 /n) | f(x) | 2 dx 

n-*a 3 

= lim 27 r ll2 J exp( — t 2 )g(2n~ ll H) dt. 

n-*a> 

Now since 1 1 1 11 exp ( — t 2 ) is summable and dominates the 
last integrand we can say 

lim J exp(— x 2 /n) \ f(x) | 2 dx = 2rg(0) 

n-+ co J 

i = 2x |1 / 111- 

We can now proceed to the main result. 

Theorem. Let/ £ Li fl L 2 , then / € Lia,n&\\} \\\ = 2ir\\f\\i. 



14 


Elements of Abstract Harmonic Analysis 


Proof 

j \j{ x) j *dx = f limexp(-z*/n) ] /(ic) J*dx 

since lmi ll -. a ,evp(— x*/n) = 1 Now since the sequence of 
functions (exp(— x*/n) | f(x) [*} is nonnegative and also 
monotone increasing so that 

supjj exp(-x*/n) |/(x) |* rfxj = Iim Jexp(-x*/ n ) |/(x) (Vx 
ne can apply Fatou’s Jemma t and say 

J 1 /(*) |*dx < lun J exp(— x*/n) 1 /(*) l 5 dx 
“ 2*- U/lll 

by Lemma 2 This proves that / € Lj, or that ( / f 1 is sum- 
mable But since )/|* is sumniable, then 1*1/}* dominates 
exp(— x*/n) l/(x) |* Therefore 

lun j e\p( — x*/») I /(x) 1* dx => j lim |/(x) )*exp( — x*/n) dx 
or 

/|/(*)|><fc-|l/l|S = 2'll/ll» QED 

Definition. Consider a function /(t) Define 

[m, ui<« 

/-(*) - 

[0, t > n 

Theorem. Let / € L*, then/, € L\ n Li and /, € Li for all 
li n t 

positive integers n and the sequence /, » a function m 

Lt f i e m the mean square sense) 


t See appendix to Chapter 1 



1 . Fourier Transform for Functions in Li 


15 


Proof. 


/ ifn(t) | dt = f \f(t) | dt 

— R 


< (/" 1/(0 r-dtj /2 (fi.dtj' 2 

< l|/||.(2n)‘« 

or /„ 6 Li. Clearly, /„ £ Z, 2 . Therefore /„ 6 Li n Z/ 2 . By the 
previous theorem 

/n € £2- 

We will now show that the sequence { /„} is a Cauchy 
sequence; i.e. 

lim || /„ - f n || 2 = 0. 

n,m-* co 

Since /„ — f m 6 L t D L 2 , we can say 

II A - ||l = 2ir ||/„ — f m ||I. 

But (assume n > m) 

[ 1 1 1 > n 

fn — fm ~ 0 if 

1 1 1 1 < m 
so 

ll/n - fm Hi = [ m \f(t)\ 2 dt + f\m | 2 dl, 

but since / 6 L 2 , each of the above terms must go the zero 
as n, m — » <». Thus { /„} is a Cauchy sequence in L 2 . Since 
Li is a complete space the sequence /„ must converge to 
another function in L 2 . This completes the proof. 

Appendix to Chapter 1 

Certain theorems from real variables will be stated here 
for the reader’s convenience. The proofs can be found in 
Natanson p]. 



16 


Elements of Abstract Harmonic Analysis 


1. Fatou's Lemma. If the sequence of measurable and non- 
negative functions /j, f 3 , • • - converges to the function F(z) 
almost everywhere on the set E, then 

j F(x) dx < supjjV»(*) dxj 

2. Lebesgue's Theorem on Dominated Convergence. Let 

a sequence of measurable functions fi(z), /»(z), •••, 
converging almost everywhere to a function fix) , be defined 
on a set E If there exists a function H[x) summabte on E 
such that for all n and x 

\fn(x)\<H{x), 

then 

Iim j fn(x) dx ~ J f(x) dx 

3. Fubini's Theorem. Let fix, y) be a summable function 
defined on the rectangle R{a < x < b, c < y < d) 

Then 

(1) the function /(x, y) considered as a function of y alone 
will be summable on £c, d] for almost all x G [a, 6], 

(2) if Q is the set of those x £ fa, 6] for which f(x, y) is 
summable on C c . dl, then the function 

g{x) = j J f(x, y) dy 

is summable on Q 

(3) the formula 

j Jf(x, y) dxdy = j dxj^ f(x, y ) dy 


is valid 



1. Fourier Transform for Functions in L i 


17 


4. Tonelli-Hobson Theorem. If one of the iterated integrals 

J dxj f(x, y) dy, J dy Jf(x, y) dx 

converges absolutely, the double integral 

fff(x, V ) dx dy 

converges absolutely also, and all three have the same value. 

4. Cauchy-Schwarz Inequality. If / (E L 2 and g 6 L 2 , then 

j I f(x)g(x) i <fej < | J | f(x) | 2 dx^J | g(x) | 2 dx . 

5. Minkowski’s Inequality. If / € L p and g £ L p , then 
(/ \f{x) + g(x) \*dxy < I fix) | ”dxy 


+ ( j | g{x) | p dx'j (p > 1). 


6. Holder’s Inequality. If / £ L p and g £ L p where p and g 
are such that 

1 + 1 = 1 
V 9 

and p> 1, then the product/g is summable and the inequality 

|/ l/(*)9(a ; ) I dx J < \ f{x) | p dxj (^J | gix) 

is valid. 


Exercises 

1. Let {/„} be a sequence of functions in Lj(— 00 , 00 ) 
such that || /„ -/||,^Oasn-> °°. Then show that 
lim n ^. aj / n (a:) = f(x) uniformly on R. 



18 


Elements of Abstract Harmonic Analysis 


2 Suppose/ € ii(— oo, «) Prove that 

lun J(x) = 0 

»*±o> 

3 Show that every point of continuity of / is a Lebsegue 
point of / 

4 If/€ L p (— <» oo ) then prove that 

hm f {f(x + A) - f(x) \*dx = 0 

A-0 •'--00 

5 Let { /„) be a sequence in L„ such that lim,-„ {{/•—/ ll» 

=» 0 and such that/* » g Then show that / =» g a e 

6 If / and g are continuous functions defined on R such 
that / ■= g a e then show that / = g 

7 Let { U\ be a sequence in L } If hm,-*, ||/» — / IU * ® 
then prove for any g Li 

hm j f„{x)g(x) dx = f /( x)g{x) dx 


References 

1 Zaanen Linear Analysis 

2 Goldberg Fourier Transforms 

3 Natansen Theory of Functions of a Real Variable 



CHAPTER 2 

The Fourier Transform on the 
Real Line for Functions in L 2 

In this chapter we will discuss the Fourier transform of 
functions in L 2 and also the inversion of such transforms. It 
will be shown that many of the results we had for Fourier 
transforms of functions in Li carry over to L 2 . Also the 
definition, several examples, and certain properties of 
Banach algebras are given. 

Fourier Transforms in t 2 

In the last section we noted that / £ L 2 implied /„ £ L x n L 2 
and that /„—>/£ L 2 and it is this limit function, f, that we 
shall take to be the Fourier transform of a function / £ L 2 . 
To paraphrase this we note that / has the property that 

lim ||/„- / 1| 2 = 0 

n-*co 

or, one writes also 

f(x ) = l.i.m. 

n-*co 

Let us pause for a moment and see how this definition of 
f(x) matches up with our old one, denoted by 

Jo = /original if / £ Ll (1 Z> 2 . 

Suppose / £ Li n L 2 so that we are assured of the existence 
of /o, then 

fo(x) = J e izt f(t) dt 

= f e ix ‘ lim/„(f) dt 

^ n-»co 

= lim J e ix, fn(t) dt 


r 


f(t)e' zl dt. 


19 



20 


Elements of Abstract Harmonic Analysis 


by noting that f Z In and applying the Lebesgue dominated 
convergence theorem Thus 

/o = Iim /. 

Although, in general, if 

9 =* 1 1 m g Mt 

this does not imply 

jr„(x) — * g(x) ae (i e pomtwise) 

we do have in this case (see Chapter 1, exercise 5) 


}•- i >t 


where 

/ «= 1 j m 

as the connection between the two modes of computing the 
transform of a function in In n L, 

It will now be shown that a result we had for / Z I* f> In 
carries over exactly to the Case where f Z In 

Theorem (Parsecal). Let / £ Lj, then 

ll/ll, » V5 1|/ 1|. 

Proof Consider the sequence { /»( By definition of f we 
have 

hm | I/. - /Hi - 0 

This implies! 

i™ II All. - 11/ II- « 


t This follows because just, as we have J| * J — 1 y ]J < ) * — V l * or 
absolute values we can show j ||/|) a — !! p |)» | < 1!/ — 9 1I» holds for 
the norm 




2. Fourier Tronsform for Functions in Lj 


21 


lim|l/„l| 2 = li/lla (2) 

ft-* co 

and, since /„ £ Li n L 2 if / £ L 2 , we can apply the theorem 
of Chapter 1 (p. 13) to get 

II /-II*- VS II/- II* (3) 

Taking the limit in (3) as n — * =° and substituting (1) in the 
left-hand side and (2) in the right-hand side gives 

11 / 11 ,-^ 11 / 11 *. 

We now turn to the problem of recovering the function 
from its transform or: 


Inversion in L 2 
Theorem. Let/, g £ L 2 . Then 

ff(x)g(x) dx = jf(x)g(x)dx. 


Remark !./,{/ £ Lj, implies /, g £ L 2 . 

Remark 2. Each of the above integrals must exist by virtue 
of Remark 1 and Holder’s inequality. 

Proof. To avoid confusion we denote by /*(£) the function 


MO 


no, i*i<fc 

0, I t I > k 


Mx) = j e ix ‘f k (t) dl 
Mx) = J dt. 



22 Element* of Abstrocl Harmonic Analysis 

(Tilts tntko sett'e bocau«c/», ff» € L\ fl L, ) It non follows 
that 

//*(»)?«.(*) df 

Sinrt the wpiv^ioii oil the right com ergev abaolntcl} we can 
applj the 1 onellv-Hobson theorem and interchange the 
order of integration to get 

y*/»(x)?«(x) dr ~ ffk(t) rt(fg.{x)t“'di 


or 

ffkix)g.{x) tlx » 1) <tt (4) 

Wt nni-t now note two farts and the revolt will be proved 
Fird it iv setn that 

hm (| v. - jr ||, « 0 =* Urn |l (/. - g ||, - 0 

beeaus*. 

II if* ~ t» lit - I! ?• - 9 1! 

from I\r^\alv theorem Second if a sequence of function? 
{A. I coimrgev to h m the norm thl> imphe. 

Inn j k y (T)k(z) sir ” j h{x)k(x) dr 

for all k(x) m L» (<ee Chapter l ext rose 7) 

With the«e two remit > m rouid ht u> take the limit as 
» -* *e of each side of (41 

lww ** & 



2. Fourier Transform for Functions in I 2 


23 


or 


) dt = Jfk(t)g(t) dt. 


It now remains only to let k become infinite. 

Before proceeding to our next result which is the inversion 
formula it is advisable to note that 

/ will mean transform f first and then take its complex 
conjugate. 

Theorem. Let / £ L 2 and let g = /, then / = (2t)~ 1 p. In 
words this says: if you start with the conjugate of a function’s 
transform and you then transform it and conjugate it, 
dividing by 2rr yields the original function. 

Proof. If we can show 


ll/-^lk = o, 

2ir 


then we will be done. Omitting arguments, we consider 



We now observe that 


/ fg = Jfg = /// = l[/[li 


= ||/ II2- 



24 


Elements of Abstract Harmonic Analysis 


Now, since II 5 115 - 2ir||g||5 - 2r||/||i ■> 4ir> ||/||1, 

11/ - ^ III - 11/115 - ll/tli - 11/115 + ~ 11/115 - o 

QED 

As a corollary we can now state the following 

Invertlon Formula. If f £ L*, then 

m = 1 1 m — * r dx 

2tr 

Proof Let f € Li and let / — g Then 

,m £? 8nd } -i s 

More precisely 

/(t) “lira ~ f e ,t( )(x) dx 
Taking the conjugate of both sides, 

f(t) = 1 i m — f e"**'/(x) dx 

n-d> 2ir 

We can now state the following result 

Theorem (Plancherel). Let / € Then there exists a func- 

tion, / € Li, such that 

/(*) = hm f e ul f(t) dt 

and 

f(0 =1 lm ~ j (T^'jix) dx, 

also 


ll/H* = (2 t)'« H/ll, 



2. Fourier Transform for Functions in 1 2 


25 


As a consequence of this inversion formula and Plancherel’s 
theorem we can now assert that every function in L 2 can be 
viewed as the transform of another function L 2 . We note 
that if two functions have the same transform they must 
be equal almost everywhere by the inversion formula and 
will be considered as actually the same function. We can 
summarize this by saying that the mapping of L 2 into L 2 
by the Fourier transform is both 1-1 and .onto. 

Normed and Banach Algebras 

Definition. A set X is called a normed algebra over C, where 
C is the field of complex numbers, if 

(1) X is a normed linear space over C, 

(2) X is a ring with respect to two internal operations, 
the additive operation being the vector addition 
in (1), 

(3) k is a member of C and x and y are vectors in X, then 

k{xy) = ( kx)y = x{ky). 

Thus we have established a link between the external and 
internal multiplications. 

(4) The norm must have the property that, with respect 
to internal or ring multiplication, 

II x v II < IMI-IMI. x >v € x. 

If, in addition, A is a Banach or B-space ( complete normed 
linear space) then X is called a Banach algebra. 

To clarify the concept of a Banach algebra several ex- 
amples will be considered. In most cases the examples will 
clearly be linear spaces and all we will do is to define a multi- 
plication and a norm that satisfies property (4). One final 
word of warning about (4) is in order though: the product 
intended on the left-hand side of the inequality is meant to 
be the ring product or product with respect to internal 
multiplication. 



26 Efemenfi of Abstract Harmonic Analysis 

Example 1. LetAf be a Banach ■space and consider L(X, X), 
the class of all bounded linear transformations on X (A 
linear transformation A on a normed linear space is said to 
be bounded if there exists a positive real number, K, such 
that 

f J A (x) (| < K JI z f| for all x m X ) 

We define multiplication of bounded linear transformations 
in the natural way (AB(x) = A (B(x) ) ) and take as a 
norm on L(X X), 


II A || - sup 


II 


Example 2 Let X Cfa 6}, complex-valued continuous 
functions on fa, 6] W e define multiplication in the point wise 
fashion and take as the norm of / € X the maximum value 
that the absolute value of / attains on the closed interval or 
|!/|| = max \/(x) I 

«uu 

Example 3 Let TT be the set of all absolutely convergent 
trigonometric «enes _ c.e tn 1 and take as the norm of 
any *(t) m It 

11 *(0 II “ 11 It c.e"' H = £ I c. 1, 

with multiplication taken as the Cauchy product 

Example 4 Let A consist of all functions analytic in the 
open unit disk in the complex plane and continuous in the 
closed unit disk We «haU take as the norm of a function, /, 
in A 

ll/ll = max|/(z) 1 = max |/(z) |, 

i«i<i i *i-i 


and multiply such functions pointwise 



27 


2. Fourier Transform for Functions in L 2 

Example 5. Let the space P n+1 denote the space of all 
polynomials of degree less than or equal to n. We must first 
define multiplication of two such polynomials in a way that 
is closed. To this end we take 


h(t)g(t) = 

t=o 

where 

n n 

C* = E a M, Hi) = E a / y > and g(t ) = 

}+l=~k j = 0 1=0 

Next, we define || aft || = E"=° I “i I- 

Example 6. Suppose Lj is the space considered in Chapter 1 
with multiplication defined as 

fg = / *g- 

We showed in Chapter 1 that 

Example 7. Suppose G = jci, o- 2 , m„} is any finite 

group. Consider the class of all complex-valued functions 
on G ; i.e. all / such that /: G —> C. This class is denoted by 
Li(G) and is called the group algebra of G 
We will denote multiplication in Li(G) by * and define 
the product of two functions/, g in Li((7) as follows: 

= E /Ws (°v)- 

ffioj—ffk 

Now, since we can write <r; = o- t cr ; - _1 , we have (f * g) (<r*) = 
E"-i/(‘ r * ff r 1 M<r/). (A careful inspection of this process 
reveals that it is, formally, similar to the usual multiplication 
of polynomials.) As a norm on Li(G) we shall take 

11/ 1! = E !/(-••) I 

i=I 



28 


Elements of Abstract Harmonic Analysis 


and we shall actually demonstrate here that this norm 
satisfies (4) Consider 

II/-SII - El </•«)(«> I 

“ Si I2/(<r*«rr I )ff(<r J ) [ 

t i 

^ H Si I 

* i 

* ]£ I ?(«■/) I 2 l/(wr*) I 
-ll/ll II All 

Example 8 Let £ denote the set of all integers, and take 
Li(Z) to be all complex valued functions, f, on Z such that 

fi W»> \ < w 

Let us define multiplication of two functions / and g from 
Li(Z) / * g as 

(/•!!)<») - E /(» - 

and take the norm of / to be 

ll/ll- Ei/wi 

Without verifying the claims we assert that this too forms a 
Banach algebra 

Examples 6 7, and 8 have many features in common 
each had a multiplication that resembled (or actually was 
as in the case of Example 6) convolution Actually through 
the use of an abstract integral we could have defined multi- 
plication in Examples 7 and 8 as convolution also by ap- 
propriately assigning a measure to the set of all subsets of 
the space The similarity extends beyond this, however 



2. Fourier Transform for Functions in I 2 


29 


The space underlying the function space was in each case a 
group but more than that each space was actually a topo- 
logical space, the discrete topology being assigned in the 
case of Examples 7 and 8. But there is still more that each 
had in common: Each is a locally compact topological group. 
We leave these observations now to pursue something else 
that is more immediate but will return to them again, later. 


Analytic Properties of Functions from C into Banach Algebras 

We now turn our attention toward mappings from the 
complex numbers, denoted by C, into Banach spaces, X, 
over C. Symbolically, if D is a region in C, then let 

x: D->X 


X — *• x(X). 


It now makes sense to make the following: 


Definition. The function, x(X), is said to be analytic in D if 


x'(Xo) = lini 

X-*Xo 


s(X) — x(Xp) 
X - X 0 


exists for all X 0 in D. Note here that the limit is taken in the 
sense of the norm on X. 

Suppose we consider the mapping 

x f 

CD D-^X-rC 


where /is a linear functional on X has the property that there 
exists a k > 0 such that 

|/(x) | < fc jjxjj forallx€X; 

/ is called a bounded linear functional on X. 

Here / is a complex-valued function of z £ A" and }x is 
just a function of the complex variable, X. Denoting the 
class of bounded linear functionals on X by X we state, 



30 


Elements of Abstract Harmon c Analys 


without proof that 


sup 

11*11*0 


l/M I 
II *11 


is actually a norm on X and we write 

l/M I 


11/ II - sup 


lull 


We also state without proof the following 

Theorem A linear functional is bounded if and only if it is 
continuous 

In view of these facts suppose 

/ X — C and / € \ 

If x(X) is analytic in D then /(*(X)) is analytic in D 
To prove this consider 

1 m /l<W) -/(*!>.)) _ A »(>) - IQ..A 

X — Xo X — Xo / 

-f(x (Xo)) 

by using linearity and the fact that boundedness is equivalent 
to continuity which allows us to interchange the order of 
computing the functional value and taking the limit M e 
will now prove Liouville s theorem for vector valued func 
tions of a complex variable 

Theorem Let x C — * X where X is a Banach space If 
i(X) is analytic m the entire plane and bounded (le 
II *(X) H < M where M > 0 for all X € C) then x(X) must 
be constant 

Proof As noted above /(x(X)) is analytic when / £ ^ 
Since/ € ^ we can say there exists a l > 0 such that 

1/(*(X))| < Ml *(X) || <kv 



2. Fourier Transform for Functions in L 2 31 

But since fx is just a complex-valued function of a complex 
variable that is also entire and bounded we can apply the 
ordinary Liouville theorem to it and assert that / must be 
constant. Let a and /3 be any two complex numbers. Then 

/(*(«)) =/(*(«) 

or, by linearity, 

f(x(a) - x(fi)) = 0. 

But / is any bounded linear functional on X. Therefore, 
since every bounded linear functional vanishes on the vector 
x(a) — x(/3), it follows from a consequence of the Hahn- 
Banach theorem, that the vector must be zero or 

x(a ) = x(P) for any a, p 

which is equivalent to saying 

x(\) — a constant for all X. 

We can now extend another familiar result. 

X 

Theorem. Let X be a Banach space and let C 3 D > X. 

If x(\) is analytic in a region bounded by a rectifiable 
Jordan arc F and continuous on T, then 

f x(X) d\ = 0. 

•'r 

Before proceeding with the proof we make the following 
two observations: 

1. The integral here is defined as a limit in the norm. 

2. The integral must exist by virtue of completeness of 
space and the continuity of x(X) on the contour itself. 

Proof. Let / € X and let y = Jrz(X) dX. 

Reasoning as in the previous theorem the following manip- 
ulation is justified if / € X : 

f(y ) = J /(z(>0) dx. 



32 


Elements of Abstract Harmonic Analysis 


But /(*(A)) is a complex-valued analytic function in the 
region and continuous on r Therefore by the Cauchy 
integral theorem 

/(«) - 0 

As before since every bounded linear functional vanishes on 
y y must be zero or 


y - j x( A) dX = 0 

The following result will not be proven but follows simply 
from the above as m the classical case 

Theorem Under the same hypothesis as the preceding 
theorem 


*(& 


1 f *(A) , 


We could also show using this that x'(*e) existing im 
plies x (X 0 ) exists and so on and also generate a power series 
expansion about X 0 for x(X) with radius of convergence 



the power series being x(X) = fX — Xo) n , where 


We now focus attention on some of the internal properties 
of Banach algebras 

Theorem Suppose X is a normed algebra then nng multi- 
plication is a continuous operation 



2, Fourier Transform for Functions in Lj 


33 


Proof. Let x , j/, x q , j/ 0 6 X. Then 

II *2/ - x 0 y 0 || = II (* - xo) ( y - yo ) + x 0 (y - yo) 

+ {x - x 0 )y 0 || 

^ II x ~ xo || || y - y 0 1| + || x 0 || || y - yo || 
+ II 2/o II II* - 2o||. 

Taking the limit as x, y — » x 0 , yo yields the desired result. 

Theorem. Let X ^ |0J be a Banach algebra with identity, 
e ; i.e., 

ex = xe = x for all x £ X. 

If x £ X and || e — x || < 1, then: 

1. x is a unit of X (i.e., x has an inverse) and 

2. 2 T 1 = e + Z"( e - *)”■ 

Remark. The summation Z”( e — x)” exists (in the norm) 
for 

II (e - x) n || < || e - x ||" < 1. 

Proof of Theorem. Write 

x — e — (e — x) 

and take 


(e — (e — x))(e + Z( e — x)") = e -f (e - x) 

1 

+ (e — x) 2 + ••• — (e — x) — (e — x) 2 — ••• 
= e. QED 


Exercise 

Let X be a Banach space. Suppose y,"_iXn converges abso- 
lutely, i.e., yi”_, || x„ || converges. Then show that 
converges, and || || < Z II *» ||. 

References 

1. Naimark, N armed Rings. 

2. Gel’fand, Raikov, and Silov, Commutative Normed Rings. Amer. 
Math. Soc. Transl., No. 5. 

3. Taylor, A. E., An Introduction to Functional Analysis. 

4. Kolmogorov and Fomin, Functional Analysis, Vol. 1. 



CHAPTER 3 


Regular Points and Spectrum 


In this chapter we continue with our study of Banach 
algebras and briefly introduce the Gel’fand theory of com- 
mutative Banach algebras 

We state and prove the following 

Theorem 1 Let X be a Banach algebra with identity 

Let X be a complex number such that |X[ > ||*||, 
then x — Xe where e is the identity of X, is a unit 

Proof First we note that if x — Xe is a unit, then Xe — * 
is also Since 


x — Xe = X(X 'x — e), 

then if we can show (X -, x — e) is a unit we will be done 
To this end consider the quantity (e — X -1 x) Taking the 
norm of e — (e — X l x) gives 



which is less than one by hypothesis Using the theorem on 
p 33 then e — X ‘x is a unit which implies (x — Xe) is a 
unit We now note that 


Xe — x 


>H) 


34 



3. Regular Points and Spectrum 


35 





= 5 >-». X "- 1 

7J — 1 


or 


(Xe — x)- 1 = '%2\-’ , x n - 1 . 

n= I 


Example 1. An immediate application of this theorem is 
available to those familiar with functional analysis: 

Let X be a complex Banach space and let L(X, X) denote 
the class of all bounded linear transformations mapping X 
into X. The set L(X, X) is a Banach algebra with identity, f 
where we take the norm of A € L(X, X) to be 


|| A !! = sup 

* 5*0 


jUM II 
11*11 ' 


By the theorem, if X is a complex number such that 


1*1 > IIAII, 

(X — A) -1 exists and (X — A)~ l = y^,”-.X~’ 1 A"" 1 . 


Let X be a Banach algebra with identity, e, and call the 
set of all units of X, U. Symbolically 

U = (a: 6 X | x exists). 


f See Taylor [1] for a discussion of this. 




36 


Elements of Abstract Harmonic Analysis 


It is noted that certainly e 6 U Consider a one-neighbor- 
hood of e, Si{e), le , 

&(e) - {*€*(||e-*||<lj 
By the theorem on p 33, 

St(e) C V 

Now, if i ^ 17, then xx~ l = e Since ring multiplication is 
continuous (see p 32) there must exist a neighborhood of x, 
N{ x) (in the norm) Buch that the set 
N(x)x ~ > C S t (a) 

where 

N(x)ir' = (l»-‘ I y ( N(x) | 

Let y € N(x) Then 

V*r' e Stic) C V 

which implies yi -1 is a unit also This means that there 
exists an element z € X such that (yx~ l )z = y{x~'z) =» «, 
and zyx~‘ = eor (ar*z)y = e 

This nnplies that y € V But y was any pomt in N(x) and 
x was any element of V so that every pomt of U is contained 
in a neighborhood that lies wholly within U, therefore U is 
an open set 

Thus the set of units, U, is an open subset of the Banach 
algebra X 

Theorem. The mapping 

/ U —* V 

given by f{x) — x~ l is continuous 

Proof All we need show to prove this is that if the sequence 
i n — * x, then I* -1 — ► i -1 

Let 1 1 «) be a sequence from U such that — » x (in the 

norm) This implies x~ x x« —* e or for any < > 0 there exists 
an X(e) such that 

ii *-■*. - « ii < « 


for n> N 



3. Regular Points and Spectrum 


37 


Choose Ni such that || x 1 x n c II ^ 1 for tt Ai and 
consider the series 

e + Y( e - a r l x n ) k for n > Ni. 

1 

By the theorem on p. 33 and the fact that || x l x n — e || < 1 
we assert that the series converges absolutely to ( x~ x x „ )-’ = 
x -ix. From the absolute convergence, then, 

|| e - x^x || < Y || (e - at-'ain) 1 ' || 


= Y !l x ~ k ( x ~ x ^ k II 

*=1 


< £ II s" 1 11*11 (*-*»)ll‘- (,) 

Jt=l 

Since x -> x n or || x - x„ || -» 0 we can choose N 2 such that 
n > N 2 will make (*) as small as desired or 

|| e - x~'x || — > 0 

which means 

x~ l x — > e or a:” 1 — > ar 1 . QED 

Before proceeding further two definitions will be stated. 


Definition 1. Let X (E C. 

If x - Xe is a unit, then X is called a regular point of x. 


Definition 2. The set of nonregular points of z is called the 
spectrum of x and will be denoted by <r(x) . Thus X x , 
then (a; — Xe) $ U. 


Example 2. To those familiar with functional ana ysis 
following fact is apparent: Let L(X, X) be t e aMC " 
algebra as in Example 1, and let A 6 L(X, A). , 

is « regnkr point of X. then (X - A)- € h(X, X) wh.oh 
implies X € p{A) where p(A) is the resolvent se o 



38 


Elements of Abstract Harmon C Analysis 


Suppose conversely that X 6 p(A) Then (X — .4) 1 
must be bounded and the range £(X — -4) must be dense 
in X Tie would like to show now that X must be a regular 
point of x and for this we ask the reader to recall the 
following two theorems from functional analysis f 

(1) If a linear transformation is bounded on all of the 
Banach space Y then it is a closed linear transformation 

a 

(2) Suppose XD D-t K If X is a Banach space and B a 
closed linear transformation such that is bounded on the 
range B(D) then B(D) is closed 

In view o f these two theorems we can say that R (X — A ) = 
R(h — A) ( the closure of R(X — A)) But ft(X — A) was 
dense in X or R(\ — A) - X Therefore 

R(\ - A) * X 

This says that (X — A) is bounded and defined on all of 
X hence (X - A) 1 € L(X X) which implies X is a regular 
point Thus we have 

p(A) - {set of regular points of A} 

Taking complements we obtain 

C(p(A)) — <r(A) {usual definition of o-(A)} 

= tr(A) {as given m Definition 2} 

Hence our old notion of spectrum (le the functional 
analysis notion) agrees with that given m Definition 2 

Compactness of the Spectrum 

Let X® be a regular point of x This means that 
xx, = (x - \fi) € V 


t See Taylor [2) 



3. Regulor Points and Spectrum 


39 


Since U is an open set, there exists a neighborhood of x Xo , 
iV(x x „), such that 

iV(x Xt ) C u. 

Consider the function y\ = x — Xe; i.e. 

2/x: C->X 

X — > x — Xe. 

This function is continuous for, certainly, 

X — ■» X 0 implies (x — Xe) — > (x — X 0 e) . 

Since it is continuous, given a neighborhood of x x „, 
iV(x Xo ) C U, 

there must exist a neighborhood of X 0 , A 7 (Ac>) C C, such that 

yx e N(x\ 0 ) c u 

for X € iV(Xo). 

Thus, y\ € U. In view of this, all the points in iV(X 0 ) are 
regular points; hence the set of regular points is an open set 
because every point has a neighborhood lying wholly within 
the set of regular points. This implies that <r(x) is a closed 
set in C. 

We already know by Theorem 1, that if | X | > || x ||, 
then (x — Xe) _I exists which means that c(x) must be con- 
tained in the closed disk of radius || x ||, or 

X 6 <t(x) => | X | < || x ||. 

Therefore o-(x) is closed and bounded. By the Heine-Borel 
theorem for the plane, then 


cr(x) is a compact set. 


Using the definition of analyticity given in Chapter 2 we 
would now like to show that the function 


x(X) = (x — Xe) -1 . 






40 


Elements of Abstract Harmome Analysis 


where X is regular is an analytic function of X over the set 
of regular points To do this we will first obtain a certain 
equality 

Consider the function 

i C*->X 

X -» (x - Xe) 1 

where C x C C is the set of regular points and x is some 
element of X (Note we are identifying the element x with 
the function x ) 

Let Xi and Xj be regular points and take 
®(Xi) *x(X*) ~ (x ~ Xje)x(Xj) 

«=((* — Xj«) + (X* - Xj)e)x(Xj) 

- (x - + (M - Xi)x(X,) 

= e + (Xj — Xi)x(Xj) 

Thus 

z(\ t ) - x(X t ) + (A* - 
or 

x(X 3 ) - x(Xj) = (Xi - XOxfXOxfX,) 

Theorem x(X) = (x — Xe) ‘ is analytic in the (open) set 
of all regular points 

Proof LetX X 0 be regular points By the preceding estimate 

Urn =, h m x(Xo)x(X) (1) 

x-x» X — Xo x-x 0 

Now we note that the function x — Xe is a continuous func 
tion of X Second it is recalled that in a Banach algebra 

x„ — * x => x„ 1 — * x~ i 



3. Regular Points and Spectrum 


41 


and third since, by continuity, 

lim x — Xe = x — Xoe, 
x-x 0 

then 

lim ( x — Xe) -1 = (x — X 0 e)~K 

X-Xo 

Hence the limit, (1), is just 

( x — Xoe) 2 . 

Example 3. With L(X, X ) as in Examples 1 and 2, let 
A £ L(X, X), X € p{A), and let R\ denote the resolvent 
operator, i.e., R\ = (X — A)~ r . Now 

|Ex = |(X-^)- 1 =-|M-X)-> = -E 2 


Theorem. Let Xbea Banach algebra with identity let x £ X. 
Then o(x) <f>. 


Proof. Suppose 
Then 


cr(x) = 0. 


x(X) = (x — Xe) -1 


is analytic throughout C. But 


e — 


x 

X 


11 II 
— > e 


ii ii 

as | X | — » oo which implies (e — (x/X)) _1 — » e -1 = e as 
| X | — x «> . Now 

|i (at - Xe)" 1 1| = | X- 1 1 || (e - X-x)- || -» 0 

as | X | -^ co (2). 

Therefore (x — Xe) -1 is a bounded, entire function. By 
the form of Liouville’s theorem proved in Chapter 2, then 

(x — Xe) -1 = a constant 



42 


Element* of Abstract Harmonic Analysis 


and by (2) above, the constant must be zero But this is 
nonsense, for if (j — Xe)" 1 = 0 how can 
(x — Xe)(x — Xe)" 1 — e ? 

Thus the assumption, o{x) — 0, has led to a contradiction 
and therefore »(x) ^ 0 QED 
We will now state and prove an important theorem about 
the structure of certain types of Banach algebras 
Theorem. If X is a complex Banach algebra with identity 
and, in addition, if X is also a division algebra, then X is 
isomorphic to C 

Proof Let x 6 X Knowing that &(x) ^ 0, let X 6 <r(x) 
Consider {x - Xe) Since X £ <r(x), (x - Xe)- 1 does not 
exist But we are in a division algebra and all elements have 
inverses except zero Hence 

(x — Xe) = 0 or x — Xe 

Now we have it, given any x f X, it is equal to 6ome scalar 
multiple of identity Thus we have the rather natural mapping 
from X — > C where the image of any x ^ Xe is taken to be 
X or 

X —> C where Xe — » X 

It is exceedingly simple to verify that 
Xie X*e — * X% Xi 

XieX 2 e — > XjXj and if ct £ C that aXe — » aX 

Further, the mapping is clearly 1-1 and onto which establishes 
the isomorphism We note that the mapping is an isometry 
if ||el| = 1 

Some new definitions are now necessary 

Definition. The real number 

r r (x) = sup [ X | 

M*) 

is sniff tn hp flip x-nertrnl rndiun of x 


3. Regular Points and Spectrum 


43 


Remark. Since | X | > || x || implies X $ <r(x), then certainly 
r,(x) < || a: ||, 

i.e. <j(x) C M where M represents the closed circle shown 
below. 



Before proceeding to our next result we will need the 
following: 

Lemma ( Special case of the spectral mapping theorem ). Let X 
be a commutative Banach algebra with identity, then 
<r(x n ) = o-(x)", where x 6 X. 

Proof. Suppose first that 

X 9^ 0 and X € <r(x n ) 

and let cui, a 2 , — , u>„ be the nth roots of X. Now 

x" — Xe = (x — cure) • • • (x — oj„e) . 

but at least one of the factors on the right must be non- 
invertible. Suppose it is (x — u>,-e) . This means that 

w, 6 <r(x) which implies cu” € a(x) n or X € c(x) n . 


Therefore 


o-(x") C <r(x) n . 


(3) 




u 


Elements of Abstract Harmonic Analysis 


On the other hand, suppose a (E <r(x) B and let ft, ft, • ■ 
ft be the nth roots of a, so that 

ft £ a(x) for some i 


Now consider 

(x — fte) ( x — fte) •••(« — fte) = x B - ae 

If x “ — ae was invertible, then we could multiply by 
(x B — ae) -1 to get 

(x„ - ae) _1 C(* - fte) • • • (x - fte) 3 *= e 
Since X is a commutative algebra we can write 

(x» - ae) -1 £(x - fte)--*(x - fte)3(i - fte) = e 
which says x — fte has an inverse Therefore 
x“ — ae is not invertible and a € <r(x") or o-(x)" C <r(x") 

(4) 

Finally if 0 6 a (x n ) , then x" is not invertible and therefore 
x is not invertible, so 0 6 v(x), whence 0 € a(x)* Con- 
versely if 0 £ <t(x) b , then 0 £ <r(x) so x is not invertible, 
consequently x B cannot be invertible so 

0 £ <r(x B ) 

Combining this with (3) and (4) we have 
<r(x“) = <r(x) B QED 

Theorem. Let x £ X (usual assumptions about X) Then 
r,(x) < hm || x B |j 1,B 



3. Regular Points and Spectrum 


45 


Proof. We first note that since <r(x n ) = <r(x)”, then 
T,{x n ) = sup I X I = sup I X I = sup I n I" = / sup I ft I Y* 

Xecr(x n ) Xe<r(x) n peff(x) l pftr(x) } ■ 

Thus 


r <r (x n ) = r,{x) n . 

\ 

Since 

r„(x n ) < 1| x n || 

we also have 

(r c (x)) n < 1| re" |l or r,(x) < || x n || 1/n 
for any n. Therefore 

r„(x) < lim || x " |l I/n . QED 

n 

Since 

lim || x " | | I/n < lim || x n || I/n 

n n 

if we can show 

t c {x) = lim || x" H l/ ", 

n 

we will have 

r,(x) = lim || x n [| 1,n . 

n-^co 

To this end consider the following: 

Theorem. Under the same assumptions about X, let x £ X. 
Then: 


(1) If | X | > r„{x), then 

(Xe — x)~ l = EX-"*- 1 . (“) 

1 

(2) If the series, (a), converges for | X | = t c (x), then it 
represents (Xe — x) -1 . 

(3) The series (a) diverges for j X | < r„(x). 



46 


Elements of Abstract Harmon c Analys : 


Proof LetX be a regular point and let ( X | > r»(x) (See 
Fig 1) 


44 



(1) By the definition of r,(x) (Xe - x) 1 must exist for 
| X | > r,(x ) By the theorem on p 40 (Xe — x) 1 is an 
analytic function throughout the set of regular points and 
therefore (Xe — x) * is analytic for | X | > r„(x) Analyticity 
however implies the function possesses a untque Laurent 
expansion Thus the expression 


~ H x* 1 

for (Ac — x) -1 for ] X | > || x || must agree with the expres- 
sion for the inverse for | X | > r.{x) or (see p 35) 

(Xe - x) 1 - £x "i" 1 i X | > r,(x) 

(2) Suppose (a) converges Then we contend that it 
represents (Xe — x) -1 for 

<>e - x) 1 - -+‘I- 1 - £X-*Z* 



3. Regular Points and Spectrum 


47 


Changing the index of summation in the first term yields 
(Xe — x ) = e -J- ^\~ n x n — ^X~ n x n 

• i i i 

= e 

which establishes (2), and also (3), namely: 

(3) If X 6 <r(x), then (a) must diverge. Suppose | X 0 1 < 
r„(x) and suppose (a) converges at X = X 0 . If (a) converges 
at all it must, by the preceding part, converge to (Xe — x) -1 . 

Since (a) is a series in 1/X, then if it converges for Xd it 
must also converge for j X | > j X 0 1. 

But there must exist Xx £ <r(x) such that | Xi | > | X 0 | and 
(a) must converge there. Thus the assumption of convergence 
at a X 0 where | X 0 | < r„(x) has led to contradictory results. 
Therefore 

(«) diverges for | X | < r„{x) 
or 


7v(x) is the radius of convergence for (a). QED 
We note that the power series in 1/X must have 
lim |J x” || 1/n 

n 

as radius of convergence bj r the Cauchy-Hadamard formula. 
Therefore 

r„(x) = lim |! x" || I/ ". 

71 

Combining this with the theorem immediately preceding 
the above theorem we have 


r„(x) = lim || x” || 1/n 

n-*co 




48 


Element* of Abjtract Harmon c Analysis 


Introduction to the Gel’fand Theory of Commutative Banach 
Algebras 

At this point we will introduce the Gelfand theory of 
commutative Banach algebras In the following discussion X 
will be assumed to be a commutative Banach algebra with 
identity e and further it will be supposed that the norm of 
e is one 

Definition A subset / of X is said to be an ideal if 

(1) J is a subspace of X and 

(2) x € X y £ / implies ry £ I 

Remark It is easily verified that the above conditions are 
equivalent to the following two conditions 

(1) x y £ 1 implies (x + y) £ / and 

(2) i € / a € X implies zx £ I 

Definition An ideal is said to be proper if there exists at least 
one element in X that is not in the ideal 

Theorem If / is a proper ideal in X then I the closure of 
/ is a proper ideal in X 

Proof First it will be shown that I is an ideal in X and 
then it will be show n that if / is proper then / is also 

(a) I is an ideal 

Let x y £ / There must exist sequences |z,} and |y*} 
such that Zn — * x and y»—*y where x. € f for all n Since 
/ is an ideal (z„ + y n ) £ I for all n Since this is true for all 
n then the limit of the sequence (x n + y»! is in I, or 

(* + y) € / 

f»ow let z £ X and let x„ — * x where x £ / Since zx* £ I 
for all n it follows that zx £ I Thus I is an ideal 

(b) / is proper 

Let z € / If x~ l exists then x~ l x £ / or e £ / Since 
multiplication by any element is allowable this would say 
that I = X Therefore if x £ / then ar 1 cannot exist In 



3. Regular Paints and Spectrum 


49 


other words proper ideals must be made up solely of singular 
elements (i.e. nonunits). 

Letting U denote the set of units of X as before then we 
now have, denoting the complement of U by CU, 


This implies 


I C CU. 
ZC CU. 


Sinc e U is an open set, CU is closed which means CU = 
CU or 

ZC CU. 

Now, since e | CU, e $ I. Thus Z is a proper ideal which is 
the desired result. 

Before proceeding another definition is necessary. 

Definition. An ideal, M, is called maximal if the only ideal 
that contains M properly is all of X, or i f J D M properly 
where J is an ideal in X, then J must equal X. 

Theorem. (1) If Z is a proper ideal, then there exists a 
maximal ideal M such that M C Z. 

(2) If x is a singular element, then there exists a maximal 
ideal that contains x. 


Proof. This result is a standard result from modern algebra 
and in it we will not have to make use of the fact that X 
is a Banach algebra, the result being true for any algebra 
with an identity. 

(1) Let S denote the set of all proper ideals which contain 
Z. It is easily verified that set inclusion induces a partial 
ordering on .S’. Let T = {Z„J be any totally ordered subset 
of S. If we can show now that T has an upper bound that 
is also a member of S, we will have shown that S is induc- 
tively ordered and can apply Zorn’s lemmaf to assert the 
existence of maximal element for S. Clearly an upper bound 
for T is just U a I a . We must now show that U a Z a is also a 


t Zom’s lemma: Every inductively ordered set has a maximal element. 



50 


Elements of Abstract Harmon c Analys 


proper ideal In general the union of ideals is not an ideal 
but since T was totally ordered it follows that U „I a is also 
an ideal and it is proper since e $ U«I B Lastly since every I a 
contains 1 it follows that U a / a is a member of S Therefore 
S is inductively ordered with respect to set inclusion and 
by Zorns lemma must have a maximal element which is 
clearly a maximal ideal Thus there is at least one maximal 
ideal that contains I 

(2) Let x be a singular element and consider the principal 
ideal 

(x) - {xy | y € X} 

Pt is simple to verify that (x) is actually an ideal 3 We note 
that e cannot belong to (x) for if it did there would have to 
exist a y € X such xy •= e which would say that x is a unit 
contrary to our assumption Further we see that ex = 
x € (x) so that (x) is nonempty Thus (x) is a proper ideal 
in A and we can apply part (1) of the theorem to assert 
that there exists some maxima] idea] containing (x) and 
hence also containing the element x QED 
We now note that any maximal ideal must be closed for 
as we noted earlier M being a proper ideal implies M is a 
proper ideal too In any case M Q M and if the inclusion 
were proper this would contradict the maximahty of M 
Therefore if M is a maximal ideal M = M 
We will now show that given any maximal ideal m X 
there is another Banach algebra immediately available 

The Quotient Algebra 

With the same assumptions about X let M be a maximal 
ideal in X It immediately follows that M is a closed sub 
space in X We now claim without proof that the quotient 
algebra with respect to M X/AI is also a Banach algebra 
commutative with identity A typical element of X/AI is 
the eoset x + M u here x is an element of X Addition of 
cosets is defined as 

(x + M) + (y + M) - (x + y) + M 



3. Regular Points and Spectrum 


51 


multiplication as 

(x + M) (y + M) = xy + M, 
and scalar multiplication as, if a is a scalar, 
a(x -f- M) — ax + M. 

As a norm on X/M we take 

\\(x + M) |! = inf || y ||. 

X/ex+AT 

It is to be noted that the faet that M is closed is essential 
here, if what we have called a norm is to be truly a norm 
for we desire the property that 

|| a: + M || > 0 

and equals 0 if and only if x is such that x + M is the zero 
element of X/M. It is here that M being closed is essential. 
For our next result we will need the following theorem from 
algebra. 

Theorem. If R is a commutative ring with identity and M a 
maximal ideal in R, then R/M is a field. 

(The reader is referred to any good book on modern algebra 
for a proof of this; e.g., Van der Waerden [230 

By this theorem then, using the fact that X is a commuta- 
tive ring with identity, we can say that X/M, in addition 
to being a Banach algebra, is also a field. By the theorem 
of p. 42 it now follows that X/M is isometrically isomorphic 
to C, the complex numbers. 

Since, by the same theorem, it follows that every element 
of X/M is some scalar multiple of identity; i.e. given x and 
M there exists a scalar x{M) such that x + M = 
x(M)(e + ill), the isomorphism can actually be exhibited 
by the following mapping 

iso 

X » X/M » c (a) 

x — * x + M = x(M) ( e -j- M) — *■ x(M). 



52 


Elements of Abstract Harmon c Analys : 


But there is something else going on here as well If we 
choose some particular element X we can map the set 
of all maximal ideals 

M the set of all maximal ideals m X into the complex 
numbers via 

x M C 
M -* x(M) 

where we have identified the element i f X with the func 
tion x M —» C 

We now claim the following assertions are true for mappings 
of this type 

Theorem 

(1) If x — y + z then x(M) - y{M) + z(Af) 

(2) If x - ay where a is a scalar then x(M) - ay{M ) 

(3) If x - yz then x(M) = y(M)z{M) 

(4) e{M) - 1 

(5) x(M) — 0 if and only if x belongs to M 

(6) If M N are distinct maximal ideals then there exists 
an x € X such that x (M) ^ x(N ) 

(7) | z(M) | <||*H 

Proof Statements (1) (4) all follow directly from the 
fact that the mapping (a) above is a homomorphism from 
X into C 

(5) x(ilf) + AT = 0 + <=>*€** 

(6) If M y* N then there exists an element x that belongs 
to M but not to N Therefore 

x(Af) = 0 but x(N) 9* 0 

(7) Write 

*-f &f- z(Sf)(e + Jff) 



By the definition of || x + M ||, 

11*11 > II* + M|| * \\x(M)(e + M) || 

= 1 x(M) | || e + M ||. (b) 

But 

|| e + M || < || e || = 1. 

Suppose the strict inequality prevailed; i.e. suppose 
|| e + M || < 1. 

This implies that there must be a t/ 6 e + M such that 

II y II < i. 


If y £ e 4- M, however, it can be written 

y — e + x for some x £ M. 

But 

y=e+x=e— ( — x). 

Now since 

1 > II 2/11 = lie - (-*) ||, 

this implies (— x) is a unit by the theorem on p. 33. This is 
impossible though because M is proper. Therefore 

II e + M || = 1.. 

Substituting this in (b) gives 

|l * || > I x(M) I || e + M || = | x(M) | 

which proves (7). 


Exercises 

1. Let X be a complex normed algebra which is also a division 
algebra. Prove that X is isomorphic to C. 

2. Let X be a Banach algebra and let x 6 X. x is called a 
topological divisor of zero if there exists a sequence [x n \ 
in X such that || x„ || = 1 and either xx n — » 0 or x n x — > 0. 



If i is a topological divisor of zero, show that x is not a 
unit 

3 Let X be a Banach algebra Show that if 0 is the only 
topological divisor of zero in X, then X is isomorphic to C 

4 Let X be a Banach algebra If there exists a constant 
m > 0 such that )) xy |) > m JJ * || )| y )| for all x, y 6 X, 
then show that X is isomorphic to C 

5 Let Y be a closed subalgebra of the Banach algebra X 
Let x € Y and denote by ax{x) and ay(x) , the spectrum 
of x in X and V, respectively Show that 

(a) <rx(x) C *r(x) 

(b) each boundary point of trr(i) is also a boundary 

point of ax{x) 

6 If X is a commutative Banach algebra and if x, y € X, 
show that r,(xy) < r,(x)r,(y) 

References 

1 Toyicv A i. An Jnlraduelttm to Fvnthevusl Analyst* 

2 Ian der Waerden Modem Algebra 


CHAPTER 4 


More on the Gel'fand Theory 
and an Introduction to 
Point Set Topology 


In this chapter we will continue our discussion of the 
Gel’fand theory of commutative Banach algebras and will 
then briefly review some aspects of point set topology that 
will be necessary in the later development. 

Throughout the following discussion X will denote a com- 
mutative Banach algebra with identity, e, where the norm 
of e is equal to one. 

Theorem 1. Let x £ X and let M € M, where M is the set 
of all maximal ideals in X. Then a; is a unit if and only if 
x{M) 0 for any M € M. 

Proof. It is first noted that a; is a unit if and only if x is 
not a member of any maximal ideal. This is justified by 
noting: 

(1) if a unit belongs to an ideal, the ideal cannot be proper, 
and 

(2) any singular element is contained in some maximal 
ideal (see p. 49) . 

Further, x{M) = 0 if and only if x £ M . Combining these 
results with part (5) of the theorem on p. 52 implies x is 
a unit if and only if x(M) 9^ 0 for any M 6 M. QED. 

Our next result yields a new way to characterize the spec- 
trum of an element. 

Theorem 2. Denote the spectrum of x by <r(x). Then 
<r(a 0 = [x(M) I M € M}. 

Proof. Let x(M) = X where M G M and consider 
(x - Ae) (M ) : 

(x - Xe) (M) = x(M) - \e(M) 

55 



56 


Elements of Abstract Harmonic Analysis 


but, since e(Af) =■ l, 

(x - Xe) (HI) = X - X « 0 

<=> (x — Xe) £ ill 

<=> (x — Xe) is not a unit 

Therefore 

X € *(x) or {x(ill) J M e AT} C <r(x) 
Conversely, suppose 

X € «r(x) 

This implies x — Xe is not a unit By Theorem 1, then, there 
must exist an M € M such that 

(x - Xe)(HI) = 0 
or 

x{M) - X = 0 
or 

x{M) = X 

Theorem 3. Let x £ X Then 

hm I) x“ || ,,B ■ sup | x(AT) | 

» AUft 

Proof. As proved in Chapter 3 
r,(x) — Inn |f xr ||"" 

= sup | X | 

= sup | x(M) | by Theorem 2 

Mtfi 

Definition. The set is called the radical of X 

Remark. Since 0 £ M for every M £ HI, then ccrt&wly 

o e n jr«ftAf 



4. Gel'fcmd Theory and Point Set Topology 


57 


Definition. If (1 mXiM = {0 j , then X is called semisimple. 
We 'now observe that 

x £ f) M <=> x(M) = 0 for all M £ M 

MeM 

<=> sup | x(M) | = 0 

Mail 

<=*■ lirn || x ” || 1M = 0. 

n-+ca 

Elements, x, with the property that there exists an n such 
that x n = 0 are called nilpotent elements, while if 

lirn || x n || ,/n = 0, 

n-» cd 

then x is called a generalized nilpotent element. 

Having defined the set M we can now speak of mapping 
it into other sets. In particular we would like to consider 
mappings of M into C and we will denote the class of all 
functions that map M into C by F{M) ; i.e., 

F(M) : all complex valued functions on M. 

Certainly, functions of the type mentioned at the end of 
Chapter 3 are in this category where we identify the elements 
of X with the function 

x: M->C 
M-^-x(M). 

Identifying the elements of X with these functions defined 
over M can be viewed as the mapping (where, in this case, 
we denote the function x by its functional value x(M)) 

X -> F(M ) 

x-*x{M). (*) 

It is clear that F(M) is actually an algebra with respect 
to the usual operations of addition and multiplication of 
functions and scalar multiplication of a function by a com- 
plex number. With this in mind it is now claimed that the 



58 


Elements of Abstract Harmonic Analysis 


last mapping, (•), is actually a homomorphism from X mto 
F(Af) This follows from the last theorem m Chapter 3, 
1 e , we claim that under (*) 

* + V -* (* + y) (M) = i (A/) + V(M) , 
if a is a scalar 

aX-> {ca){M) = ax(M) 

and 

W -* (xy) (Jlf ) = x(M)y{M) 

Having realized that (•) is a homomorphism one might 
next inquire about the kernel of (•) If x is a member of the 
kernel then x(M) = 0 for all M £ M This is equivalent to 
saying that a: is a member of every M £ M, or that x belongs 
to the radical of X If the kernel consists only of the zero 
element however then the mapping (*) must be 1-1 or is 
actually an isomorphism We can summarize these results 
by the following 

Theorem If X is senusimple then 

X F[St) (•) 

x — » x(A/) 

is an isomorphism 

Before proceeding further a concrete example of these 
results will be given 

Example As commutative Banach algebra with identity 
take 

w = Jx(t) = £ ‘W" I i, I I < °°| 

where 


11*11 - SW-1 

(See Example 3 m Chapter 2 for some discussion of W ) 



4. Gel'fand Theory and Point Set Topology 


59 


First we would like to determine what the class of maximal 
ideals is in W. 

Let xo = e xl and let xo(M ) = a. Now consider xj4 = e~ il . 
By the above-mentioned homomorphism it follows that 
Xq 1 (M) — or 1 . Using part (7) of the last theorem proved 
in Chapter 3 we now have the following two equations: 

f«l = I MM) | < || x„ || = 1 
I a" 1 1 = I M(M) | < H I) = 1 

which implies 

M = i. 

Since | a | = 1, then there must exist a U £ £0, 2v) such 
that cc = e*‘°. At this point let us consider the mapping (») 
for W: 

iso 

g: W —* W/M — > C 


x 

> x(M) 





iint 

y £171 fft 


and ultimately 

CO CO 

x(<) = Ecne- - ]C c " ei "'° = x(M). (1) 

— CD — 03 

Let us examine this last statement in more detail. Cer- 
tainly, since g is a homomorphism, for any finite sum, 
Sn £ W, Sn = 2-tfC„e <n( we have Sn(M) = >jLvc„e''"°. 
But we also have | x(M) | < || x (| in general which, in this 
case, reduces to 

| S n (M) - x{M) | < || Sn - x ||. 

Since the last term on the right must go to zero as N 
becomes infinite, we have established continuity of g in 
this case which justifies (1) 4 Thus, to each maximal ideal M 

f Actually the mapping on p. 51, of which g is a special case, is always 
continuous by part (7) of the theorem on p. 52. 



60 


Elements of Abstract Harmonic Analysis 


is associated a point t a € f0, 2i r) such that M consists of 
those x € W which vanish at to (see part (5) of the theorem 
on p 52) 

Conversely, let U € C0> 2») and consider the set 

m =. e x | £«•■» - o) 

It is clear that if is an ideal m IV We would now like to 
show that it is maximal Suppose I was an ideal in W such 
that 7 3 if properly Then there must be a y £ I such that 
y <£ M Since y $ M, y(to) ^ 0 Now for any 2 6 W we can 
write 

1(0 = m vW + H " iw» <0 ] 

But the first member on the right is in I w hile the second 
vanishes at f<> and is therefore in M This implies that z € I 
But z was any element of W hence I = IF Let us now 
review what we have done First we saw that given any if 
there corresponded a U € fO 2ir) Now we have that given 
any t 0 € f0 2r ) we can construct a maximal ideal Hence 
there exists a one to-one correspondence between the set if 
of maximal ideals of }V and the real numbers in the interval 
fO 2 it) or M <-* [0 2jt) Consider an x(t) = S-» Cb€ "" 
such that it is nonvamshmg for every t This implies x $ if 
for any if € if Tins says that x(t) is a unit or that 
1/a: ( <) £ W This result was first proved by Wiener in an 
entirely different context and is stated below 
Theorem (Wiener) If is absolutely convergent 

and vanishes nowhere, then (l/V.^^e'" 1 ) can be expanded 
in an absolutely convergent trigonometric series 

Topology 

This discussion is intended only as a brief introduction, for 
more detailed information the reader is referred to the books 
by Bourbaki flj or Kelley [2] 



4 . Gel'fand Theory and Point Set Topology 


61 


A Topological Space 

Let X be a set and let 0 be a collection of subsets of X. 
If the collection of subsets satisfies 

(1) U a O a € 0, a € A, where each 0„ € 6 and A is an 
arbitrary indexing set, 

( 2 ) n" =1 o,- e o, n < », o, e 6, 

(3) X, 0 € 0, 

then the pair (X, 0) is called a topological space. Usually the 
members of 0 are called open sets, and 0 is called a topology 
for X. Frequently one speaks of just X as being a topological 
space. 


Examples of Topological Spaces 

Example 1. X an arbitrary set and 0 = (0, X}. This is 
called the trivial topology. It is easy to verify that this is 
indeed a topological space. 

Example 2. X an arbitrary set and 0 = P(X) where 
P(X) is the collection of all subsets of' X and is usually 
called the power set. This is usually called the discrete topology 
because every subset, even discrete points, are open sets in 
this topology. 

Example 3. Let X be euclidean n-space and let 0 be the 
collection of open sets as they are usually defined in euclidean 
n-space. 

Example 4. Let X be an infinite set and let 0 consist of 
0, X, and any subset A whose complement CA is finite. We 
will actually verify that this is a topological space. 

(1) C( = D a C(A a ), where A a 6 6 , which is 

clearly finite. Hence 6 is closed with respect to arbitrary 
unions. 

(2) Let A h Ag, •••, An € 6. The intersection' 

(fLiU, 6 6 , since C( rfLiXi) = U^UCd.,-, is finite.^ 

(3) This requirement is satisfied by the way 0 was 
defined. 



62 Elements of Abstract Harmonic Analysis 

Further Topological Notions 

In the following definitions ( X , 0) is assumed to be a 
topological space 

Definition 1. A set F is said to be closed if the complement 
of F is open, i e , CF € 6 

Definition 2. An element x € X is called an adherence point 
of a subset E C X if every open set containing x contains a 
point of E 

Remark It immediately follows from Definition 2 that 
every element of £ is an adherence point of E 

Definition 3. The set of all adherence points of a subset E, 
denoted by L, is called the closure of E By virtue of the 
remark following Definition 2 we have E <Z E 

Definition 4 Let X and Y be two topological spaces The 
function / X — » 1 is said to be continuous if for any open 
set, 0, in r, / ‘(0) is an open set in A r 

Definition 5. Let X and Y be two topological spaces The 
function / X — * Y is said to be a homeomorphtsm if 

(1) the mapping is both 1-1 and onto, and 

(2) / is continuous and / 1 is continuous or, to summarize 
the second requirement, / is bicontinvovs 

We note that a homeomorphism, since it preserves open sets, 
preserves all topological properties, abstractly the two sets 
are indistinguishable topologically 

We now proceed to the intuitively plausible result that the 
composite function of two continuous functions is itself 
continuous 

Theorem 1. Let X, Y, and Z be topological spaces and let 
the functions / and g be continuous functions 


X-> Y —* Z 



4. Gel'fond Theory and Point Set Topology 63 

Then the composite function h(x) = g(f(x)) is also 
continuous. 

Proof. Let 0 be an open set in Z. Then 
h- l (0) =/- I (r‘(0)). 

Since g is continuous, g~ l (0) is an open set in Y. Since / is 
continuous, f~ 1 (g~ 1 (0)) is an open set in X which completes 
the proof. 

Before proceeding to our next result we state, without 
proof, the following: 

Theorem 2. If A C B, then A C B. 

The proof of this statement is quite short and the reader 
should verify it for himself. 

Theorem 3. Let ( X , 0) b e a topological space and let 
E\, E? C X. Then Ei U IE — E, u La- 

Proof. Certainly 

Ei C Ei l) Ez 


which implies, by Theorem 2, 


K\ C U $2- 

(2) 

By the same reasoning, we also have 


E 2 CZ El U E 2 . 

(3) 

Combining (2) and (3) we obtain 



Ki U Kn C Ei U Ez. 


Now let x be any element in E\ U £•>. This means, by the 
definition of closure, that any open set, 0, containing x 
must also contain a point of Ei U Ei. Now there are two 
possibilities: either 

(1) every open set containing x contains points of E\ 
which would imply x £ £\ which means x £ Ei U Ei, or 

(2) there exists an open set Oi containing x such that 
Oi n Ei = 0. Suppose this is the case and let 0 2 be any open 
set containing x. Now we have x £ 0 ifl 0 : . 



64 


Element] of Abstract Harmon c Analysis 


Since Oi fi 02 is an open set containing x we must have 
(0! fl Ot) fl (£, U Ei) ^ 0 

But since 

0i n Ei => 0 

we must have 


(Oi n 0 2 ) n E, ?f 0 


which implies 

Oi fl Ei 5 ^ 0 or x € £j 

and hence 


a € £i u £, 


Theorem 4 2 =■ £ 

Proof By the remark following Definition 3 we have 

£CE 


Thus all we need show is that 

Ic£ 

To this end fet x £ 2? and let 0 be any open set containing 
x Thug 

O n £ * 0 

or there must exist ay 6 0 such that y 6 £ Since y 6 £ 
and y £ 0 0 fl £ ^ 0 Thus any open set containing a (: £ 
must also contain points of E Hence 

a € £ or 2? C £ 

Combining this with £ C 2? we obtain 
£ - Z? QED 


The next theorem will give us a more useful analytic way of 
describing a closed set 

Theorem 5 £ is closed if and only if E = £ 



4. Gel'fand Theory and Point Set Topology 


65 


Proof ( Necessity ) . Suppose E is closed. By definition, CE is 
open. Let x € CE. Since CE ft E = 0, x $ E. Since no point 
of E can be in CE, we must have E C E. But, in general, 
E C E. Hence E = E. 

( Sufficiency ). Suppose E = £ and let x € CE. Certainly 
x cannot belong to E because E = E. Hence there must be 
at least one open set 0 containing x that has nothing in 
common with E or, for any x in CE, there must exist an 
open set 0(x) such that 0(x) fl E = 0. Thus 

CE C U 0{x). 

xzCE 

Since no 0(x) has anything in common with E we have 
CE = U 0{x). 

xcCE 

Since we are able to write CE as a union of open sets, CE 
is open. Therefore E is closed. QED 
The next theorem yields another very important property 
of continuous functions; namely, that they map closures into 
closures. 

Theorem 6. Let X and Y be two topological spaces and let 
/: X -f Y. 

Then f is continuous if and only if for every set E C X 

nE) cm. 

Proof ( Necessity ) . Suppose/is continuous, and let E be any 
subset of X. Let x 6 E. Now take any open set 0 in Y con- 
taining f(x). f(x) € 0 implies x 6 / -1 (0) which must be 
an open set since / is continuous. Since x € E, however, 
/ _1 (0) fl E 0 which impl ies 0 fl f(E) ^ 0. Therefore 
f(x) 6 f(E) or f(E) C f(E), which proves that the condi- 
tion is necessary. 

( Sufficiency ). Suppose the condition is satisfied and let 
0 2 be open in F. Let U = / -I (0 2 ). 



66 


Elements of Abstract Harmonic Analysi 


We next note that f(CU) — f(X) n COt The condition 

implies 

f(CU) C /(A) n COi 

Let x € CU and suppose x€Ulix£U~ then 

f(x) € Oj But we also have 

f(x) € J(X) f) CO, 

which implies 

Ot n (/(X) n C0 2 ) * 0 

which is clearly impossible because 0 2 n COj => 0 Hence 
x $ V orx € CU which impl ies C U C CU Since CU C CU, 
m general, w e have CU = CU which, by Theorem 5, im- 
plies CU is closed Therefore U is open which demonstrates 
that the condition is sufficient QED 

We now proceed to an alternative way of topologmng a 
space, called 

The Neighborhood Approach 

Let X be a set and let x € X Then the collection of sub- 
sets of X, denoted by V(x), are called neighborhoods of x, if 
they satisfy the following four conditions 

1 Suppose U € V(x) Then if N D U, N e V(x) In 
words, this states that if U is a neighborhood of x, then all 
supersets of U must also be neighborhoods of x 

2 If U € V(x), then x £ U 

3 V (x) is closed with respect to finite intersections, le, 
if U lt U t , --•,£/„£ r(z), then P^U, € V(x) 

4 Let V g F(x) Then there must exist a set If € V(x) 
such that for all y m 17, V 6 V{y) Loosely phrased this 
says that if I' is a neighborhood of z, then it is also a neigh- 
borhood of all points "sufficiently close" to x 

Having defined these neighborhoods we can now define 
an "open" set in terms of these neighborhoods The use of 
the word "open” will be justified by showing that the “open 
sets so defined do indeed satisfy the three axioms for a 
topological space 



4. Gel'fand Theory and Point Set Topology 


67 


Definition 6. A set 0 C X is called open if for any x £ 0 
there exists an N (x) 6 V (x) such that x £ N(x') C 0. To 
paraphrase this one might say that a set is open if it is a 
neighborhood of each of its points. 

We will now show that these open sets do give a topology 
for X. 

Clearly X and 0 are open according to this definition. 

1. Arbitrary uniom. Denote by 0 a (where a ranges 
through some index set A ) the collection of open sets defined 
by the above definition and let 

w = uo„ 

a 

where a ranges through B C A. If x £ W, then, for some a, 
x 6 0*. This implies there exists an N(x) £ V (x) such that 
x € N(x) C O a C W. Therefore W is open according to 
this definition. 

2. Finite intersections. Closure will be demonstrated for 
an intersection of two open sets from which it immediately 
follows, by induction, for any finite number. Let Oi, 0 2 . be 
open sets, and let x £ Ox n 0 2 : 

x £ Ox => x € Ni(x) C 0 1 

x £ 0 2 =» x 6 lV 2 (x) C 0 2 . 

Hence 

x 6 (JVi(s) n N 2 (x)) C (Oi n 0 2 ). 

By property 3 of neighborhoods it follows that Ni{x) PI 
IV, (i) is a neighborhood of x. Hence Cfi fl 0 2 is an open set. 

In summary, we have seen the following process take 
place: Given neighborhoods, V(x), they immediately give 
rise to a class of sets, 0, that satisfy the axioms for a topo- 
logical space. 

Diagrammatically, 

X, F(x) 

\ 

6 



6S 


Element* of Abttracf Harmonic Analysis 


We -will now show that this picture can be extended We 
will now start with a topological space and define a class of 
sets V (re) from the open sets that will be shown to satisfy 
the neighborhood axioms This will give us the following 
diagram 

X, V(x) 

N 

0 

\ 

V(x) 

It will turn out that the classes of sets V (x) and f (x) are 
actually the same 

Definition 7. Let ( X , 6) be a topological space U will be 
called a neighborhood of x if there exists an open set 0 £ 0 
such that x £ 0 and 0 C U Loosely phrased this says that 
a neighborhood of x must be "big enough to fit an open set 
about the point x The set of neighborhoods of x will be 
denoted by ^(x) 

first it will be shown that the class of sets defined by 
Definition 7 satisfy the four neighborhood axioms 
Axiom i Let U £ V'(x) and suppose V D U Since 
U £ V(x) there exists an open set 0 £ 6, such that x £ 0 
and U 0 which implies V 3 0 

Axiom 2 It is clear from Definition 7 that if V £ ^(#)> 
then x £ V 

Axiom 3 Let V» and W be neighborhoods of x Then we 
have open sets, Oj Oj £ 0 such that 

x £ 0. C Vi, x € OtCVt 

which implies 

x £ O x (\0,CV ifl V t 

Thus Vj n 1 j is a neighborhood and we have established 
that the neighborhoods defined by Definition 7 are closed 
with respect to finite intersections 



■4. Gel'fand Theory and Point Set Topology 69 

Axiom A:. Let V £ V ( x ) . Then there must exist an open set 
Ox 6 0 such that x £ 0 X and 0 X C V. But an open set is a 
neighborhood of any of its points by Definition 7 for, given 
any z £ 0, where 0 6 0, certainly 26 0 C 0. Hence given 
any y £ 0 X 

y £ 0 X c V 

which implies F 6 V(y). QED We now wish to show that 
the following diagram is valid: 



where the arrows pointing down should be read as “gives 
rise to,” and the arrow going up means that V (x) = V(x). 

Suppose now that we are given (X, V(x)). Using Defini- 
tion 6 we can construct a topological space {-X, 0) ; Applying 
Definition 7 to (X, 6) we can then define ( X , V(x)). We 
will show V(x) — V (x) . 

Suppose the above-mentioned process has been performed 
and let 

U e V(x). 

There must exist an open set, O x , such that 
x 6 O x C U. 

Since O x is open according to Definition 6 , however, there 
must exist a neighborhood of x, N(x) 6 V (x) , such that 

x e N(x) co.cc/. 

Hence V is a superset of N{x) which implies V 6 V (x) . 
Thus 

V(x) C V(x). 



70 


dements of Abifracf Harmonic Analysis 


Now let F 6 V(z) and let 

o- !»n r € vm 


Certainly x € O and, by the way we have defined 0, 


Hence 


y € 0=* F6 V(y) => y 6 F 

oc f 


Now we must show that 0 is open according to Definition 6 
Let y be any point in 0 As noted before, V 6 V(y) By 
Axiom 4, pertaining to neighborhoods there must exist 
IF € V(y) such that for all z 6 TF, V 6 F(z) Hence IF C 0 
But y was any point in 0 and we have found a neighborhood 
of y contained m 0 which implies that 0 is open by Defim 
tion 6 Thus V € £(x), and combining this with our preuous 
result, we have (x) = F(x) We now wish to demonstrate 
the validity of the following situation (with the same inter- 
pretation of the arrows as in the previous diagram) 



i e , we wish to show 0 = 0’ 

Proof Let O 6 6 and let x € O We must find an N(x) € 
F(ar) such that x € N(x) C O Since 0 6 F(x), we have 
* € 0 C O which implies 0 6 0' 

Conversely, let 0 6 0' Then x 6 O implies the existence 
of N(x) 6 F(x) such that 


x 6 N{x) C 0 

which implies, by Definition 7, that there is a set 0, 6 0 



4. Ce l'fand Theory and Point Set Topology 


71 


such that 

x6 0 X C N(x) C 0. 
Hence 0 can be written 

0 - U 0 X 

xeO 

which imphes 0 £ 0. QED 


Exercises 

1. Let X be a commutative Banach algebra with identity. 
Let R be the radical of X. Show that X/R is semisimple. 

2. Let Xi and X be two topological spaces. Prove that a 
1-1 onto mapping /: Xi — > X is a homeomorphism if 
and only if f(A) = /(A) for all A C X- 

3. Construct an example to show that the closure opera- 
tion is not necessarily preserved by a continuous mapping. 

4. Let X be a topological space and let E C X Prove that 
E is the smallest closed set containing E. 

5. Let (X, 6) be a topological space and let E C X. x 6 E 
is called an interior point of E if there exists an 0 £ 0 
such that x 6 0 C E. Let E° denote the set of all interior 
points of E. Prove the following: 

(a) (E n F)° = F° n F° (b) CE = ( CE )° 

(c) C(E° ) = CE (d) (E°)° = E° 

(e) E a is the largest open set contained in E. 

CO CO 

6. Give an example for which U X ^ U 

i=l i-l 

7. Give examples for which 


(a) E n F E n F (b) (E U F)° ^ U F° 

8. Let X — Z, the set of integers, and let p be a fixed prime. 
V will be called a neighborhood of n if U contains all 



72 Element* of Abstract Harmonic Analys s 

n + wp* for some k and all m — 0, ±1, ±2, • Show 
that the neighborhood axioms are satisfied 

/ 

9 Let Xi and X 2 be two topological spaces, and X 2 —* X 2 
Show that / if continuous if the inverse image of every 
closed set is closed and conversely 

10 Let Xj and X* be two topological spaces and Xi — * Xi 
Show that / is continuous if for each x £ Xi and any 
neighborhood Vs of /(x) there exists a neighborhood Ft 
of x such that /(Ft) C F 2 and conversely 

References 

1 Bourbaki T opologte General* 

The reader might consult this book for a detailed discussion of the 
neighborhood approach 

2 Kelley General Topology 



CHAPTER 5 


Further Topological Notions 


In this chapter we will pursue further the topological no- 
tions introduced in Chapter 4, and terminate the study of 
topology, for its own sake. Finally we will apply the topo- 
logical notions to the set of all maximal ideals of a Banach 
algebra. 

Bases, Fundamental Systems of Neighborhoods, and 
Subbases 

The class of all open sets, 0, in a given space can be a 
very large class indeed. If possible, given a class of open sets, 
0, we would like to cut down the number of sets we must 
.focus attention on and concentrate on a smaller class of sets. 

Definition U Let (X, 0) be a given topological space and 
consider a collection of open sets \B a \. This collection is 
said to form a basis for the topology if every open set from 0 
can be written as a union of sets in { B a ) - 

Example 1 . Consider the euclidean plane with the usual 
open sets as the topological space. The class of all open 
circles, open rectangles, open crescents, etc., are all bases 
for the topology. 

Definition 2. A collection of open sets, {B a }, in a topological 
space, (A, 0) , is said to form a basis at the point x if for every 
open set containing x there is some member of {B a j con- 
taining x and contained in the open set. Symbolically, if 
x 6 0 where 0 is an open set, there must exist a set B x G \B a ] 
such that x 6 B z C 0. 

Definition 3. A topological space is said to satisfy the first 
axiom of countability if there exists a countable basis at every 
point. 

73 


N 



74 Hemenft of Abttraci Hormone Anolyi t 

Example 2 Suppose X is a metric space with the family of 
spherical neighborhoods as the (neighborhood) topology At 
any point of the space then it is readily seen that the class 
of spherical neighborhoods with x as the center and rational 
radii forms a countable basis at the point Thus any metric 
space satisfies the first axiom of countability 

Definition 4 A topological space is said to satisfy the second 
axiom of countabiltty if there exists a countable basis for the 
topology 

Definition 5 Let ( X 1 (x)) be the topological space where 
the class of sets 1 (i) are neighborhoods A collection of 
neighborhoods B(x) of x is called & fundamental system of 
neighborhoods of xd for each neighborhood of x N{x) £ V (x) 
there exists a neighborhood B € B(x) such that B Q N(z) 

Theorem I Let X be an arbitrary set and consider a class 
of subsets of X B(x) such that at any point x 6 X the 
following three axioms are satisfied 

(BI) If V t Vi 6 B(x) then there must exist a set Vt € B{x) 
such that C Vi 0 V t 

(B2) If V £ B(x) there exists a set V € B( x) such that 
t C V and if y £ V there must exist a member of B(y) 
contained in V 

(B3) Each member of B(x ) contains x 
Then there is a topology for X such that at any point 
x € X the class of sets B(x) forms a fundamental system of 
neighborhoods of x 

Proof We define a class of sets V (x) as follows A set 
AT £ V(x) if and only if there is a set B £ B{x) such that 
B C. N We will now prove that the class V(x) forms a 
neighborhood topology for A Hence we must show that 
the four axioms Cot a collection, of seta to be called neighbor 
hoods are satisfied (see p 66) 



5. Further Topological Notions 


75 


1. Supersets. Suppose N 6 F(x) . This implies that there 
exists a V such that N D V € B{x). Certainly any set IF 
that contains N also contains V. Therefore W 6 V (x) . 

2. The element must be in all its neighborhoods. Suppose 
N 6 F(x). Then N D F € B(x). By B3 though, x £ F, 
which implies x € Ah 

3. Closed with respect to finite intersections. It suffices to 
demonstrate that the intersection of any two sets in F(x) 
is also in F(x). To this end let N\, Ah 6 F(x). Then 

2Vi D Fj € B(x) and jV 2 D F, <= B(x) 
which implies 

Ah n Ah D Fi n F 2 

which, by Bl, must contain a set F 3 £ £(x), or N x n N 2 D 
Fj 6 B(x). Therefore Ni n Ah € F(x). 

4. A neighborhood of x is a neighborhood of all points 
“sufficiently close” to x. Note that as yet we have not made 
use of B2.. It-is here that we shall need it. 

Suppose N 6 F(x) . This implies there exists a F such that 
A T 3 F 6 B(x). By B2 we have a set F' such that 

N D F D V e B(x) 

and if y 6 V', there must be a set B £ B(y) such that B C 
V' C N. Hence V' € V(y) and for all y € F', N € V(y). 
This completes the proof. 

We would now like to focus attention on an even smaller 
class of sets than the sets in a basis for a topology. 

Definition 6. A collection of open sets in a topological space 
is said to form a subbase ( subbasis ) for the topology if the 
collection of all finite intersections of sets from this class 
forms a basis for the topology. 

Example 3. Consider the real line, R, with the open 
intervals as the class of basis ^ets; i.e., all sets of the form 
(a, 6) . A subbase for this topology would be the class of all 
sets of the form (a, « ) , (—«,&). 



76 


Elements of Abstract Harmonic Analyst 


Theorem 2. Let X be any set and let S be any collection of 
subsets of X Then there exists a topology for X in which S 
is a subbase 

Proof To simplify matters we mtroduce the following 
notation 

B the class of all finite intersections of sets in S 
O the class of all unions of sets m B 

1 fl, X 6 6 To assure this we adopt the following con- 
vention, vacuous unions represent the null set and vacuous 
intersections represent the entire space t If the reader finds 
this convention distasteful, the alternative is to just put the 
sets 0 and A in 6 separately 

2 Arbitrary unions It is clear that arbitrary unions must 
be m 6 by the very definition of 6 

3 Finite intersections We will show that the intersection 
of any two sets in 6 is also m 6, the extension to any finite 
number then follows immediately by induction First it will 
be shown that B is closed with respect to finite intersections 
Let Bi Bs € B Then 

5i = (\S , Bt = ns . 

where each of the S, S, 6 S, 

Bi n B, = ( ns *) n ( nSl ) 

= Si n n S, n 5i n • n St ^ B 


t To render this intuitively plausible one notes that taking the UD, ®“ 
of fewer and fewer seta leads to fewer and fewer points being contain 
in that union whereas taking the intersection of decreasing numbers o 
sets leads to more and more points being contained in the intersection 



5 . Further Topological Notions 


77 


Now let 0 = Oi fl O 2 where 0 1; 0 2 € 6. Then 


Oi n Oi 



where {B al \, {B a ,\ are families of sets in B. Hence 
0i n 0 2 = U U(B ai n B„ 2 ) . 

ttl 02 


But since B is closed with respect- to finite intersections, 
Oi fl 02 has been written as a union of sets in B. 

We shall often refer to this topology for X, given S, as the 
topology generated, by S in a way analogous to the way one 
defines the space spanned by a set of vectors in a vector 
space. Just as in the case of the space spanned by a set of 
vectors, that this space is the smallest subspace containing 
the set of vectors, and that this subspace was actually equal 
to the intersection of all subspaces containing the set, we 
note that it is clear that the topology generated by S is 
equal to the intersection of all topologies containing S. 

Definition 7. Consider a set X and two topologies of open 
sets for X, 0\, and 0 2 . [The spaces (X, 0i) and (X, 0 2 ) 
represent two possibly different topological spaces.] 0i is said 
to be weaker than 0 2 or 0 2 is stronger than Oi, denoted by 
Ox < O 2 , if Oi C O 2 . Suppose X is any set and the collection 
{X a } is a family of topological spaces and j f a ] is a class of 
functions such that f a : X — > X a . We wish to assign a topology 
to X in such a way that each of the/„ is continuous. We note 
certainly the discrete topology assigned to X will do the 
job, albeit a rather crude one. A somewhat more refined 
choice consists of taking the intersection of all those topol- 
ogies for X with respect to which all the /„ are continuous. 
This topology is called the weak topology associated with the 
f a and can also be arrived at by taking the class of sets 
( fa* 1 (Of ) } , where O a is an open subset of X a as a subbase, S; 
i.e., take the topology generated by S. 



76 Element* of Abstract Harmonic Analysis 

The Relative Topology and Product Spaces 

Definition 8, Let (X, 0) be a topological space and let S 
be any subset of X The set S together with the collection of 
sets {0 fl £ | 0 6 0} is said to be the subspace (relahic, 
induced) topology for S 

Of course it remains to verify that the collection of sets 
{£ n O j O € 0} does indeed satisfy the axioms for a topo- 
logical space, and we shall prove this now 

1 Since 0 =* 0 n S and S «» £ n X, it is clear that S, 0 
are in the above-mentioned class 

2 Since 

(Oi n S) n (0, nS) = (Oi n o,) n S, 

closure with respect to finite intersections follows 

3 Finally, since 

(1(0. n S) - (^UO.'jnS, 

it follows that an arbitrary union of sets of the above class 
is also just the intersection of an open set from 6 and S 
which proves that (S, {£ n 0 1 0 € 6}) is a topological 
space and justifies Definition 8 

Next we will consider the cartesian product of two topo- 
logical spaces and will assign a topology to this space The 
approach used will work for the cartesian product of any 
finite number of topological spaces 

Let (Xi, Oi) and (X*, (5 t ) be two topological spaces and 
consider Xj X X* = { (ij, x,) | Xi 6 Xi, x* 6 X*J We wish to 
topologixe this space, hence we must propose some definition 
of open set in the product space 

A set U C Xt X Xj will be called an open set in the product 
space if and only if for any y € V there exists a set 0i X 0i 
such that y € Oj X Oi C 0 where 0i 6 Oi and 0i 6 Ot It 
now remains to \enfy that the collection of sets so defined 
does indeed give a topology to the product space 



5. Further Topological Notions 


79 


1. It is clear that 0 and Xi X X 2 are open according to 
the above definition. 

2. Finite intersections. Suppose U and V are open in the 
product space. Then if y £ U n V, we have 

y € U => y € Oi X 0 2 C U (a) 

and 

y 6 V => y e 0[ X 0; C F. (b) 

Combining (a) and (b) we obtain 

y £ (OiX 0 2 ) n (0[ x 00 C n F 

or 

ye (Ox n 00 x (0 2 n 0 2 ) C O n 7. 

Hence 17 n V is open according to the above definition. 

3. Arbitrary Unions. Suppose { } is a collection of open 
sets in the product space. Then if y e U a U a , there must 
exist an a such that y e U a . This implies 

y e 0 X X 0 2 C U a C U U a 

which completes the proof. 

Note: Henceforth all subsets and cartesian products, when 
considered as topological spaces, will be assumed to be 
topologized by the above topologies. 

Separation Axioms and Compactness 

To motivate the ensuing discussion consider a topological 
space with the trivial topology assigned to it. 

In a general topological space we will say a sequence [x„] 
converges to a; if every open set containing x contains almost 
all of the x„. If the trivial topology has been assigned, how- 
ever, then we see that any sequence at all converges to any 
and every point in the space. Thus the notion of convergence 
in this space becomes totally uninteresting because the limit 



80 Elements of Abstract Hormowc Analysis 

is so completely ambiguous The difficulty with the above 
problem is that there is not enough “separation” between the 
poults of the space To avoid this we shall further refine the 
class of all topological spaces by defining some separation 
axioms 

Axiom T t. Given a topological space where any two points, 
x and y, have the property that there exist open sets, 0, 
and 0 V , where x € O x and y 6 0„, such that 0* n O t = 0, 
we shall say the space satisfies Axiom T 2 A synonym for this 
is saying that the space ts Hausdorff 

Thus given any two points m a T* space we can separate 
them by enclosing them in nonmtersecting open sets Also, 
in a Tj space the limit of a sequence ji»J is unique if it exists 

Example 4 As an example of a space that is not Hausdorff 
consider the following X is any infinite set and the open sets 
are the null set, X, and all those sets whose complement is 
finite (In Chapter 4, Example 4, it is verified that this 
actually rs a topological space ) In this space it is impos- 
sible to obtain any two nonmtersecting open sets, hence 
this cannot be a Hausdorff space One can, however, given 
any two points x and y, find an open set containing x and 
not containing y Clearly the open set X — {t/i will do the 
job Hence, even though we do not have the strong separation 
imposed by the Hausdorff axiom, we do have some separation, 
and spaces which do obey this law of separation are classified 
by the following 

Axiom T t . If, given any two distinct points of a given 
topological space x and y, one can find an open set con- 
taining x but not y and also an open set containing y but 
not x, then the space is said to satisfy Axiom Ti 

Example 5 The following is an example of a space that is 
not Ti Let X = [0, 1) and let 6 - {0, [0, o) where 0 < 
a < 1 ) Without verifying that this is a topological space 
we note that given two points x and y where y > x we can 
find an open set containing x but not y but cannot find an 



5. Further Topological Notions 


81 


open set containing y that does not contain x, which brings 
us to our next separation axiom; namely: 

Axiom T 0 . A topological space with the property that for 
any two distinct points there exists an open set containing 
one but not the other is said to satisfy axiom T 0 . 

The following theorem reveals an equivalent way of saying 
that a space is Ti. 

Theorem 3. A topological space, X, is Ti if and only if every 
one point set is closed. 

Proof (Necessity) . Let X satisfy Axiom Ti, and consider 
two distinct points x and y. Certainly y 6 X — (x). Since 
X is Ti though, there exists an open set, 0„, such that y £ 0 V 
and x $ 0„. Now consider 

U O v = X — {xj. 

ycX—lzl 

Hence X — {xj is an open set and the one point set (x) is 
therefore closed. 

(Sufficiency). Suppose for every x £ X that (xj = {x}. 
If x 5^ y, then y £ X — {x} which is an open set and x £ X — 
[y\ which is also open. Thus the space is Ti. 

Definition 9. Let A be an indexing set and let (X, 0) be a 
topological space. Further, suppose {(?„), where a runs 
through A, has the property that X C U a <7 a . Then {<?„} is 
said to be a covering of X. If the set A is finite, then it is 
said to be a finite covering of X. If each G a is open, the cover- 
ing is said to be an open covering of X. 

Definition 10. A topological space is said to be compact if, 
from every open covering, one can select a finite subcovering. 

We note that the Heine-Borel theorem for the real line 
is just the statement that closed and bounded sets there, 
with respect to the basis consisting of open intervals, are 
the compact sets. 



82 


Elements of Abstract Harmonic Anolysls 


Theorem 4 A compact subspace of a Hausdorft space » 
closed 

tively, such that 0, »». ' s) Slnce s 

P'^^ t -°UU?o"Xc^ T«.e m »^nS 

0 ,J such that & - n ; m y n g v,as not 

rr ,1 u* rr — n ,11, , Suppose now that r ' |0 

mil This would m«n tint V n (0„ n B) ^ 
® Lire E 7 , 0 (0„ n Q * 0 which > »«"£ 
cause of the way the U, were chosen m the firstpta 
we have S C S C S whrch implies S - « 

Theorem 5 A closed subspace of a compact space is «®P«‘ 

Proof Let F be a closed subspace and considerj-he 
ing covering for F Let F - U 0 L„ where * , ope a 
the 0. are open Hence the entire spa “ * Lust cover X 
covering 10.1 u CF A finite number of these m Q _ 

because X is compact denote these by i 
Since they cover X we must also have 

F = U(O.nf) 

which completes the proof 

Theorem 6 The continuous image of a ,nd 

compact Symbolically if / is continuous A, is comp 
x, is a topological space then if 

/ Xi-*X t 

j(Xi) is compact 

Proof Consider any open covering of /(A'i) ” j J 
This implies X, = Since X , , is compact » ^ 

a fimte number of these must also do the job, deno 



by Ui, U t , U Thus 


Z x = U /-(f/,) =»/(Zx) =/( u /-’(f/,)) C Ut/,, 

Theorem 7. Let Xi be a compact topological space and X 2 
be a Hausdorff space and let / be a continuous 1-1 onto 
mapping of X\ onto X 2 . Then / must be a homeomorphism. 

Proof. Let E be a closed set in X 2 . By Theorem 5, E must 
be compact, and by Theorem 6 ,f(E) must also be compact. 
By Theorem 4, f(E) must be closed. Hence closed sets must 
map into closed sets under/, and since it is already 1-1, onto, 
and continuous, we have the desired homeomorphism. QED 
When one works with sets there is always a certain duality 
present; i.e., for every statement about open sets there is a 
corresponding statement about closed sets. Compactness is 
no exception, for if S is a compact space we have 

n 

U O a = S => there exists U O ai — S 

a i=l 

and the corresponding dual statement is 

n 

nC„ = 0 => there exists D C„,- = 0 

a t= 1 

where the C a are closed, which is equivalent to: 

n 

if no fl C a> . = 0, then flGL 0 . 

1=1 a 

These ideas are now summarized in the following definition 
and theorem. 

Definition 11. A family of sets in a topological space with 
the property that the intersection of any finite number of 
them is not null is said to satisfy th e finite intersection properly . 

Theorem 8. A topological space, X, is compact if and only 
if for any family, {F a }, of closed sets satisfying the finite 
intersection property, fl„F a ^ 0. 



84 


Elements of Abstract Harmonic Analysts 


Proof ( Necessity ). Suppose X is a compact and let I/*,} be 
a collection of closed sets satisfying the finite intersection 
property Suppose non that 

= 0 => u cf. = x 

Since CF„ are open sets and X is compact, we have 
UCF t « X=» C\F X = 0 

which contradicts the finite intersection property Therefore 
ft F„ ^ 0 

( Sufficiency ) Suppose the condition is satisfied and 
simultaneously that X is not compact If X is not compact, 
however, there must exist an open covering of X, (0.), with 
the property that, forever yl, 0. x U 0*, U- • • U 0„ t X which 
implies CO at D CO., n • • • n C0 ak & 0 Hence the class of sets 
{CO.} satisfies the finite intersection property and, by 
hypothesis, this implies ft.CO. s* 0 But the {0„} were a 
covering so that ft.C 0. *= 0 Hence we have arrived at a 
contradiction and X must be compact 

Next we will briefly introduce the Tychonoff topology on 
a product space and state the Tychonoff theorem 

The Tychonoff Theorem and Locally Compacf Spaces 

Consider a collection of topological spaces (X„j where 
« € A and let O a denote the open sets in X„ IV e define the 
cartesian product of {X.}, HX a , as all functions 

x A — 4 UX. 


such that x(a) € X, 

A projection mapping is defined as follows 

pr* tlx. -* X„ 
x-* *(a) 



5. Further Topological Notions 


85 


A topology for UX a ^ desired in such a way that projection 
mappings are continuous functions. We state without proof 
that a basis element for such a topology is Jl0„ such that 
0 a € 0 a and almost all (i.e. ail but a finite number) the 
0 a = X a . This topology for the product space is called the 
Tychonoff topology is the weakest topology for which all 
pr„ are continuous. 

Theorem 9. ( Tychonoff ). The product space II A*, with 
respect to the Tychonoff topology, is compact if and only if 
each of the X a is compact. 

For proof of this result, the reader is referred to the books 
of Bourbaki Q] and Kelley [J2]- 

Definition 12. A topological space, X, is called locally com- 
pact if for each x £ X there is an open set O t such that 
x £ 0 X and O z is compact. 

We note immediately that, for any n, euclidean n-spaee 
with respect to the usual topology is locally compact. 

Theorem 10. Let Xi and X 2 be topological spaces such that 
Xi is compact. Let F be a closed set in Xi X X 2 . Then the 
projection. of F on X 2 is closed. 

Proof. Let E be the projection of F on X 2 and let yo £ E. 
Thus any open set 0 containing yo has the property that 
0 n E 0. 

Consider now the set 

Go b = \x | (x, y) 6 F, y € Op, y 0 £ 0p\ 
and note that for any n 

Goi n Goi n • * • n Go n x 0. 

To verify that this is so we note that 

n 

Go 1 n Ojn ••• n a, C n Go,- 
>~i 

For if xc G 0l n 0 , n „. n 0 „ then there must exist a y € 0, n 
O 2 n- • • n 0 n such that (x, y) € F. It is further noted that 
Go, n o t n- n o„ cannot be null for any n because yo £ E and 



86 Elements of Abstract Harmonic Analysis 

y o 13 in every Of which implies Oj fl 0* n* • • n 0, n E ?£ 0 Now 
we can say that the collection of sets {0 o ,} satisfies the finite 
intersection hypothesis in a compact space Hence there 
must exist x» € fV?o, 0 which implies! (x p , y 0 ) £ F = F 
Therefore yt> € E and we have the fact that E is closed QED 
Theorem 11. Let Xi and X 2 be topological spaces and F 
and 0 be closed and open sets contained in Ai X A* Let E 
be a compact space m Xt Then the set 

U {y | (*, y ) 6 is closed (1) 

*<£ 

and the set 

fi ly I (*. y) € OJ = B is open (2) 

«r 

Proof (1) We note that 

U{y| (x,y) € F\ 

«K 

is just the projection of 

(S X X t ) n F 

on Xt All we need do now is note that F is compact and 
apply the preceding theorem 

(2) Consider U„ £ {y ( (x, y) € C0\ -A By the pre- 
ceding theorem A must be closed But the complement of 

A is just B and, hence, B is open 

Theorem 12. Let X% X 2 and Xt be topological spaces and 
let / be a continuous function / Xi X X 2 — > X } Let E be a 
compact set in A'i and 0 } be an open set in A* Then V = 
€ 0% for all (simultaneously) x 6 E] is open 
Proof Let 17 = ( 0 3 ) Since / is continuous this must 

be an open set m A'i X Xt Now we can write 

U - n [y I (x, y) € W openj 
«r 

and spply the preceding theorem to conclude that U is ope n 


f See exercise 7 



5. Further Topological Notions 


87 


A Neighborhood Topology for the Set of Maximal Ideals 
over a Banach Algebra 


Suppose X is a commutative Banach algebra with identity. 
Denote the set of all maximal ideals in X by M . We wish to 
topologize this set now and we will do this by constructing a 
class of sets satisfying Axioms Bl, B2, and B3 mentioned at 
the beginning of this chapter; i.e., we will construct a class 
of sets which will ultimately form a fundamental system of 
neighborhoods about any point in M. 

Let e be an arbitrary positive number and let x u x 2 , • • • , 
x„ be arbitrary members of X. Now let M 0 6 M. We shall 
show that the class of sets 


F(M 0 ; *i, X 2 , x„, «), M£M 

such that 


for 


| x k (M) - x*(Afo) | < € 
k — 1,2, • • • , n 


satisfies Axioms Bl, B2, and B3. 

Since M a itself is certainly in V (Mo] X\, ••*, x n , e), it is 
clear that B3 is satisfied. With regard to Bl consider the 
following: Let 

Vi = F(Af 0 ; xi, x 2 , • • •> x„, «i) 

V 2 = V (M 0 ', yi, y 2 , • • • , yk, a) ■ 

It is now clear that 


Vi nFiDF (Mo) x h • • • , x„, y h • • • , y k , min(e x , e 2 ) ) . 

We will now show that a much stronger condition than 
B2 is satisfied. We will show that any member of this class 
of sets is actually a neighborhood of every one of its points 
and, hence, in the open set topology associated with the 
neighborhood topology arising from this fundamental system, 
the sets in this fundamental system of neighborhoods are 
actually open sets. 



68 


Elements of Abstract Harmonic Analysis 


Suppose Mi € V(M 0 , x u • ■ x„, t) and suppose for every 
Jc = 3,2, • • ■, n that — xt(M q ) In this case, then, 

we must have 

V(M t , x u ••*,*.,«) C V(M 0 , zt, - in, «). 

On the other hand, suppose that x*(Mj) ^ x*(Af 0 ) for some 
k In this case consider 

6* = t Xk^Mx) ~ x k (M q ) j < c 

and take 

5 = max St 
* 

Certainly 5 < « or « — 5 > G Since t — 6 > 0, there must 
exist a real number e € (0, « — S) Consider V (Mi, x t, 
i n , c) We shall show that 

V(Mi, X t , • * *, In, c) C V(M<j, I|, • • ■, Xn, «) 
Suppose M € V(Mi, ii, •••, i«, c) Then 
| x*(M) - Xk(Mt) | 

= | Xt(M) - z k (Mi) + x t (Mi) - i*(il/«) | 

< | x t (M) - xtiMy) j -f ( x k (Mi) ~ xM) I 

< c + l 


<«— 5 + 3 = * for any 1*1, "•,» 

Therefore 

Af € y(Af.,u, •••,*„€) 

Hence given any member of this class of sets we see that it is 
a “neighborhood” of each of its points and, comparing this 
to Axiom E2, one sees that this is stronger than what is 
required 

Lastly we will show that the space AiT with the topology 
indicated above is a Hausdorff space, i c , we must show 
that for any two distinct points there exist disjoint open sets 
contamm^ the points 



5. Further Topological Notions 


89 


Let the two distinct points be Mi and M 2 . If Mi M 2 , 
then there exists an a; £ X such that a; (Mi) X x(M 2 ). [See 
part (6) of the last theorem in Chapter 3-3 Since x(Mi) ^ 
x(M 2 ) there is some real number e such that | a: (Mi) — 
x{Mi) 1 > e > 0. We now claim V(M\, x, e/2) ft V(M 2 ; 
x, e/2) = 0 to complete the proof. Suppose the above inter- 
section was not null. This would mean that there was some 
element, M, that was in both of the above sets, or that 

| x(M) - *(Afi) | < e/2 and | a ;(M) - a ;(M») | < e/2 

which, by the triangle inequality, implies 

| *(Mi) - x(M t ) | 

< | x(Mi) - x(M) | + | x(M) - x(M t ) | 
e e 

< 2 + 2 = £ 

which is contradictory to the way x was chosen. Hence 

V(Mi;x, e/2) and V (Mi; x, e/2) 

are two nonintersecting open sets containing the distinct 
points Mi and M 2 . 


Exercises 

1. Let Xi and X 2 be topological spaces and Xi X X 2 the 
product space. Prove that the mappings pr t : Ii X I 2 -> 
Xi and pr 2 : I ( X I 2 -> X 2 defined by pri(a;i, a: 2 ) = a:i 
and pr 2 (a: l , x 2 ) = x 2 , where Xi 6 Xi and x 2 £ X 2 , are 
continuous. 

2. Prove that any subspace of a Hausdorff space is also a 
Hausdorff space. 

3. Prove that the topological product of Hausdorff spaces 
is a Hausdorff space. 



90 Elements of Abstract Harmonic Analysis 

i If X is a Hausdorff space with respect to the topology Oi 
and if X is compact with respect to the topology Oi and 
if, further, Oi < 0 ly prove that 61 = 6* 

/ » 

5 Let Xi —* Xj where / and g are continuous, Xi a topo- 
logical space, Xs a Hausdorff space Prove that the set 
{* | f(x) = g(x ) { is closed 

6 Prove that if X is a Hausdorff space and if Ei and Ei 
are disjoint compact sets in X, then there exist open sets 
Oi and 0% such that Oi D E\, 0 3 Z) E t and Oi 0 Ot = 0 

7 Referring to the proof of Theorem 10, show that 

(xo l/o) € P 


References 

1 Bourbaki Topologie Gentro.lt 

2 Kelley General Topology 



CHAPTER 6 


Compactness of the Space 
of Maxima! Ideals over a 
Banach Algebra; an Introduction 
to Topological Groups 
and Star Algebras 


In this chapter we will apply just about everything covered 
so far about Banach algebras and topology to proving that 
the space of maximal ideals of a Banach algebra, com- 
mutative with identity, is compact with respect to the 
topology assigned to it in Chapter 5. After this, the notion 
of a topological group and several properties of topological 
groups will be discussed. 

Theorem. Let A" be a commutative Banach algebra with 
identity, e, and let M be the space of all maximal ideals in 
X. Then, with respect to the topology for M defined in 
Chapter 5, M is compact. 

Proof. The line of reasoning that will be employed here 
consists of making statements such as “if we can prove M 
has some property, then the proof that M is compact will 
follow”; i.e., the problem will be successively reduced to 
different problems which, if they can be proven, will yield 
the proof. 

Let x be any element of X and consider the closed circle 
in the complex plane of radius || x j| which we will denote 
by S x . By the Heine-Borel theorem for the plane, <S X is com- 
pact for every x. By the Tyehonoff theorem (Theorem 9, 
Chapter 5) , the product space, n&. where x ranges through 
X, is compact. Denote the product space, XI S x , by S. A 
typical element in S is then the function 

{ax}iex where a x £ S x . 

95 



92 


Elements of Abstract Harmonic Analysis 


A basis element in the (Tychonoff) topology for S is then a 
set of the form 

IT (a®, Xi, x 2 • • • , t) 

where o° € S, xi, it, • • • , x» are arbitrary members of X, 
and t is an arbitrary positive number,consisting of all a € S 
such that 

| a„ — ( < * 

i i < . 


I «* - «5J < « 

where aj, is the projection of a® on S xl etc Note that there 
can be only a finite number of these inequalities because 
almost all of the projections of any basis set in S must be 
the whole spare S, Consider now the mapping 

Si -*■ S /M a fixed element of Af\ 

M —* [z(M) j*tx V a: ranges through X / 

Since [by part (7) of the last theorem in Chapter 3] 
| x(M) \ < || * || we can say x(M) € S , 

We further note that the abose mapping is 1-1, for given 
any M\ y* M t there exists an 2 * € X such that xt(Mt) y* 
xo(Mt) Denoting the images of Mi and Mt by {a, I and 
1/3,1, respectively we have y* )3, 0 Thus the mapping is 
1-1 

Consider the 1 1 onto mapping of M onto its image Si C S 
l\e now claim that this is a homeomorphism Consider the 
basis set in Si 

W (or®, Xi * , x n t) 

This has as its inverse image those M € M such that 
j x*(M) — a xi } < t for k = 1, • n 
Since the mapping is onto St, there must exist an Mu 6 M 



6. Compactness of Space of Maximal Ideals 


93 


such that a 0 = \x(M<i) } xeA - which implies a% = x k (M 0 ) ■ 
The inverse image can now be written, using the notation 
introduced in Chapter 5, as 

V (M 0 ; %b x n , e) 

which is a fundamental neighborhood in M.\ Hence the 
mapping is continuous 

It is clear that the mapping takes basis elements in M 
into basis elements in Si. 

Hence we have established the desired homeomorphism. 
We now make the following: 

Contention. Si is closed. 

Before proceeding to the proof of this contention let us 
note what immediately follows from it, assuming it to be 
true: Since S is compact, if Si is closed, then Si must also 
be compact. (A closed subset of a compact space is compact.) 
If Si is compact, then, since M is homeomorphie to Si, M 
must be compact also. 

In view of this let us pursue proving that Si is closed. To 
prove this we must show that any adherence point of Si 
must actually belong to Si. Let a 0 = {a°} 6 Si. To show 
a 0 £ Si we must show that there exists an Mo such that 
al = x(Mo) for all x £ X. This then will imply that a 0 is an 
image point under the mapping M —> S and, by the definition 
of Si, will mean that a 0 belongs to Si. 

Before proceeding, three theorems, one from linear algebra 
and two from modern algebra, will be stated. 

Theorem 1. A nontrivial (not identically zero) linear func- 
tional is always an onto mapping of the vector space onto the 
underlying field. 

Theorem 2 ( Canonical theorem on homomorphism ). Let X 
and 7 be algebras and let h be a homomorphism mapping X 

onto 

onto 7; h: X — > 7. Call the kernel of the homomorphism K. 

t Showing that basis elements in <S'i have, as their inverse image, 
basis elements in M is certainly a sufficient condition for continuity. 



94 Elements of Abstract Harmonic Analys i 

Then the mapping shown below from A /A onto Y is an 
isomorphism 

Y — » X/K Y 

*-* (x + K) —h( i) 

Theorem 3 Let AT be a commutative algebra with identity 
and let Af be an ideal in X Then X/M is a field if and only 
if M is a maximal ideal 

Proceeding with our proof again consider the following 
mapping 

Y^C (1) 


It is claimed that this is a homomorphism and also that the 
mapping is onto and not identically 2 ero i e we claim the 
following There is some x € A for w hieh 

«!^o 

and if |5 € C then for any x 

cr®, = pal aif» = aj -f aj and — ojaj 

Assummg these claims to be true for the moment we have 
a nontrivial linear functional mapping Y onto C Hence we 
can apply Theorem 2 above to establish an isomorphism 
between X/M v and C where M 0 is the kernel of the original 
homomorphism (I) le we have 

1 — ► AT/V, — * C 

X — » x Mo —* a» 

Since C is a field however we can now apply Theorem 3 
above to conclude that Vo must be a maximal ideal Further 
more since under this latter isomorphism e + V* — * I 
«*{< + A/o) — * aj we must have al(« + A/#) — x + A/« 53 
*(V»)e + A/o If a] - x(A/«) * 0 we could multiplj by 



6. Compactness of Space of Maximal Ideals 


95 


its inverse to conclude e £ M 0 which is contradictory. 
Therefore a x — x(M 0 ) which was the desired result. Now it 
follows that Si is closed, therefore compact because S is, 
and finally, since M is the homeomorphic image of a compact 
set, M must itself be compact. 

Let us now return to proving the claims made about the 
mapping (1) . First we will show that it is nontrivial. Suppose 
that a° ss 0. Since a 0 £ Si though, for any x\, xz, • • • , x n £ X 
and any e > 0 there must be some element M £ M such that 

| x k (M) - <4 | = | x k (M) - 0 | = | x t (M) | < <= 

for fc = 1, 2, •••, n. (2) 

In particular, this must also be true for the identity element 
e. Hence (2) implies 

| e(M) | < e for any e > 0. 

But this is impossible, for by the last theorem in Chapter 3 
we have | e(M) | = 1. Hence the assumption aJsO has led 
to a contradiction and we can conclude that the mapping 
(1) is nontrivial. 

To prove that (1) is a homomorphism is a straightforward, 
although somewhat arduous, task. Since the proofs are all 
in the same spirit, only the fact that the mapping preserves 
products will be proven here, scalar multiplication and addi- 
tion being left as an exercise for the reader. To this end con- 
sider the following neighborhood of a?: 

TF(a°, x, y, xy, e). 

Since a 0 € Si, there must exist an M £ M such that the 
following three inequalities hold: 

| X(M) -al \ <e 

j y(M) - c£ | < e 

| (xy) (M) -«“ v | = | x(M)y (M) - <4 I < <• 



96 Elements of Abstract Harmonic Analysis 

Consider now 

! < “ I 

= | at, - + z(M)(y(M) - o£) + aj(l(jlf) - aj) J 

<!«!,- x(M)y(M) | + | x(M) 1 1 y(M) - | 

+ -„t| 

< < + « ii * ii + 1 «; i « 

because | i(Jlf) | < || * )) Now since e > 0 is arbitrary the 
result follows 

Theorem. If X is a complex algebra without identity, then 
X can be extended to an algebra with identity, X 

Proof Denote by Jt the set of all pairs 

(a, x) where a 6 C and x d X 

Define scalar multiplication, addition, and multiplication of 
elements from X as follows 

0(a, x) = (0a, 0x) where 0 € C 

( ai , Xi) -f (as, x t ) = (at + «s, + Xs) 

(ai, *i) («i. x») = (otiaj, aix 2 + a 2 Xi + XiXj) 

It is easily verified that X, with respect to these operations, 
is an algebra It is also easy to verify, by just substituting 
into the last expression, that (1, 0) is an identity for ft, and 
that the following mapping is an isomorphism 

X-X 

x -♦ (0, x ) 

Loosely speaking, we will actually make an identification 
between x and (0, x) and wnte 

fa,*) = a(l,0) -f (0, x) 

= ac + i where e *= (1, 0) 



6. Compactness of Space of Maximal Ideals 


97 


Hence we represent X as {ae + x |“ c c |) to obtain the required 
extension. 

Using the same notation as in the preceding theorem, 
suppose that X is a normed algebra without identity. If we 
define || ae + x || = | a | + || x ||, then X becomes a normed 
algebra also. If, in addition to being a normed algebra, X is 
a Banach algebra, then X will be a Banach algebra too and 
this we shall prove. 

Let {a„e + x n ] be a Cauchy sequence in X; i.e., for any 
e > 0, there exists an N such that n, m > N implies 
|| a n e + x n — (a m e + x m ) || < e. This is the same as 

(| — a n | + || x n — x n ||) < « 

which implies {«*„} is a Cauchy sequence and that {a; n } is a 
Cauchy sequence. Since C is complete, {«„} must converge 
to some af C, and, since X is a Banach algebra, { x„ } must 
converge to some x € X. Hence 

<x n e + X* y ae -f x. 

Therefore X is a Banach algebra. 

Star Algebras 

Definition. Let X be a complex algebra. X is called an 
algebra with involution (star algebra, symmetric algebra ) if 
there exists a mapping 

X->X 
x — » x* 

such that 

(1) (ax + py)* = ax* + By* (a, B € C) 

(2) a:** = x 

(3) ( xy ) * = y*x*. 

Listed below are some examples of algebras with involution. 

Example 1. Consider the class of all continuo us fu nctions 
over the closed interval [a, 6] and take x*(t) = x(t ). 



98 


Element] of Abstract Harmonic Analysis 


Example 2 Let X be a Hilbert space and consider L(X, X) , 
the class of all bounded linear transformations on X As 
noted previously, this is an algebra Let A € L{X, X) and 
take A* to be the adjoint operator of A 

Example 3 Consider W the class of all absolutely con- 
vergent tngometnc series, i e , 

ft where £ | c» j < w 

Let x £ W and take x*(t) =* £ L.e"' 


Topological Groups 

Definition A set G is said to be a topological group if and 
only if G is a topological space and 

(G1) G is a group 

(G2) G is a Hausdorff space 

(G3) the mapping 


G —> G 

g — ff -1 

is continuous 

(G4) The mapping (denoting the group operation as multi- 
plication) G X G — * G 

(gi ffs) gi?2 

is continuous 

It is to be noted that the linkage between the topological 
and group structures is provided by Axioms G3 and G4 In 
the axioms and in the ensuing discussion the prototype group 
operation wilt be taken to be multiplication and several 
examples of topological groups are listed below 



6. Compactness of Space of Maximal Ideals 


99 


Examples of Topological Groups 

Example 1. Let G be any abstract group with the discrete 
topology assigned to it. 

Example 2. The real and complex numbers with respect 
to ordinary addition and the usual topologies. 

Example 3. R* and C* (the real and complex numbers 
with zero deleted) as a multiplicative group with the usual ■ 
topologies assigned. 

Example 4. M n (R ) : the class of all n X n matrices with 
real entries, and M„(C ) : the class of all n X n matrices with 
complex entries. As the group operation we shall take addi- 
tion of matrices. To topologize the above closses we introduce 
the following distance function for either case. 

Let A = (ay) and B — (0y) be matrices with either real 
or complex entries and take the distance between A and B 
to be 


d(A, B) = max | a,-, - /9y j. 

if) 

Now, since we have defined a metric on the space, we can 
take the topology induced by considering the “spherical” 
neighborhoods as basis elements. 

Example 5. GL„(R ) : the general linear group of n X n 
matrices (nonsingular) with real entries, and GL n ( C) : the 
general linear group of n X n matrices (nonsingular) with 
complex entries. As the group operation we will take ordinary 
multiplication of matrices. Noting that GL n {R ) C M„{R) 
and GL n (C) C M n (C) we shall take the topology to be the 
induced topology from the bigger space in each case. 

Example 6. U n (C ): ( the unimodular group) all complex 
n X n matrices with determinant equal to one. We will 
take the group operation to be multiplication of matrices 
and noting that C7„(C) C GL n (C) we will assign the induced 
topology from <?L„(C) to U n {G). 



100 


Elements of AbsJroeJ Hormon c Anolyj : 


Example 7 The class of all complex valued continuous 
functions on a compact metric space X We shall take the 
law of composition to be ordinary addition For a topology 
we assign sup«x {/(*) j as norm to this space and will take 
the spherical neighborhoods so defined to be the basis ele- 
ments for an open set topology 

Several observations that immediately follow from the 
definition of a topological group will now be listed 

1 Continuity of multiplication (Axiom G4) is equivalent 
to the following condition 

Let gig* =* g% Then for any neighborhood Vt of g 3 
there exist neighborhoods Vj of pi and Vi of g t such that 
Fil sC Vi where 

T iV, *= [xy | x € V, y >E F,} 

2 Continuity of taking inverses (Axiom G3) is equivalent 
to the following condition 

For any neighborhood V of g~ l there exists a neigh 
borhood V of g such that XJ 1 C F where 

U ' - (ar-M x € V\ 

3 Statements 1 and 2 (simultaneously) are equivalent to 

Let g 3 = Qiffs 1 and V 3 be a neighborhood of <?j Then 
there exist neighborhoods of gi and g 3 V\ and \ , such that 

V\Vt 1 C fa 

Proof It will now be shown that (1) and (2) imply the 
above condition 

Let g 3 = g 3 gi 1 and let 1 3 be a neighborhood of g 3 By (1) 
there exist neighborhoods I 2 and Ut such that Q\ € Ii 
fft 1 E Ut and V 3 Ui C li By (2) now since U t 13 a neigh 
borhood of g t 1 there must be a neighborhood T * of Qi such 
that 1 f 1 C Ui Hence ViF, 1 C L which completes the 
proof that the above condition is necessary for (!) and (2) 



6. Compactness of Space of Maximal Ideals 


101 


Sufficiency. We first show that the condition implies (2). 
Write eg _I = g~ l and let W be a neighborhood of g~ l . The 
condition implies that there exists a neighborhood, U, of g 
and a neighborhood, T 7 , of e such that 

vu-' C W. 

Since e £ F, U~ l C VU~ l . Hence 
U~ ! C W 

which proves that the condition implies (2). 

Next, we show that the condition implies ( 1 ). Let = 
g s , or Let V 3 be an arbitrary neighborhood of 

g 3 . Then, by statement ( 3 ), there exist neighborhoods Fi 
and Ui of g h g: 1 , respectively, such that 

TW c Vs. 

Now since Vi is a neighborhood of g 2 \ by what has already 
been shown, we have that there exists a neighborhood F 2 
of g 2 such that V 2 l C U s , or F 2 C Thus 

ViVi C C Fa- 

The following two statements follow directly from the 
preceding. 

4 . Taking the product of any three elements of G is a 
continuous operation; i.e., let gig 2 g 3 = g* and let F4 be a 
neighborhood of gt . Then there exist neighborhoods of <71, gi, 
and r/a, Fx, Vi, and V 3 , such that ViV 2 V 3 C F<. 

• 5 . Let .<7j (72 = g 3 and let V 3 be a neighborhood of g 3 . Then 
there exist neighborhoods of gi and Qi, Vi and F 2 , such that 

V\V 2 c f 3 . 

These results will now be absorbed into the following 
observation. 

6. Let 

&g k V---g’n , ‘ = 9 



102 


Elements of Abstract Harmonic Analysis 


and V be a neighborhood of g Then there exist neighbor- 
hoods of g it g 2 , •• •, and g n , V t , V t , • • and V„ such that 

• v*' c V 

7 The following mappings are all homeomorphisms 
(1) G-+G (2) G~+G (3) (?—►£? 

9 -* 9°9 9 -> 99» 9~* r l , 

where is a fixed element of G 

Proof Since the proofs of these three statements are all 
about the same, only (2) n ill be proved here, the proofs of 
the rest being simple to prove following the same lines of 
reasoning 

(2) is 1-1 Suppose g,go = g z go This immediately implies 
ffi = after multiplying on the right by g 7 l (2) ts onto 
For any g 6 G it is clear that gg 7 x will map into it under (2) 

(2) is continuous Let h * ggc, and let V be a neighborhood 
of h By the very first observation there exist neighborhoods 
of g and go U and IF such that UW C F which implie* 
Ugo C V which establishes the continuity of (2) 

(2) is open (takes open sets into open sets) Since we can 
write the inverse function explicitly as 
(2)~ J <7— ♦ G 

9~* 997\ 

th ^rixist neig 

iis is exactly the same as what was just proved 
is a ho meomorphism 

Proof ^ p ^ a c | osec i set m G and let 0 be an open set in 
above con snJ e i cmmt ol G Then 
Let pi 

there ex g«F, Fgc and F~ l are all closed 
barbo^er d Vts an open set m G and E is any set in G, then 
that | UE, EU, and U~‘ 

proo i ire ill open 


6. Compactness of Space of Maximal Ideals 


103 


Proof. We will prove here that Fg<> is closed and that U E 
is open to illustrate how the proofs run and leave the rest 
as an exercise for the reader. 

Fg 0 is closed. Consider/: G —* G where /(g) = gg 0 , g 0 fixed. 
By 7, / is a homeomorphism. Therefore, since F is closed, 
/(F) = Fgo is closed. 

UE is open. Since U is open, /([/) = Ug 0 must also be 
open. Now we can write 

UE = U(Uffo); 

i.e., as a union of open sets. Hence UE is open. 

9. Any topological group, G, is homogeneous ; i.e., for any 
g h go € G there exists a homeomorphism of G onto G that 
takes gi into g 2 . To prove this all we need consider is the 
mapping 


/: (?-><? 

9 997*92 

and note that, since g/'g 2 is fixed, this mapping, by 7, is a 
homeomorphism. 

As to the importance of this result consider the following 
situations: 

9.1. Suppose we wished to show that G is locally compact. 
We desire for any g £ G an open set containing g such that 
the closure of the open set is compact. If we can show, just 
for the identity element, say, that there is an open set con- 
taining it whose closure is compact, we can apply 9 and con- 
clude that this is true for every element in the space. 

9.2. Suppose we wished to show that G had the discrete 
topology. If we could show that just the set (e) was open 
this would, by 9, immediately imply that this was true for 
every element in the space. 



1 0-4 Elementi of Abstract Harmonic Analysts 

10 Suppose V is a neighborhood of e Then there exists 
s neighborhood U of e such that U Q V and V = U~* In 
words, this states that within any neighborhood of the 
identity there is a symmetric neighborhood of the identity 

Proof Let 

U = V fl V~ l 

It is now clear that U is a neighborhood of the identity and 
that U C V It remains for us to show that U is symmetric 
Before proceeding to that we note that this proof will apply 
to open sets as well as neighborhoods because if 0 is an open 
set such that 

V2)0, K->DO-», 

then 

V n D 0 n O- 1 

Let a € (! This implies o £ V and a 6 V~ l , but 
a € V a -1 € F" J and a 6 F -1 =* a~ l € F 

Hence a" 1 € F ft F' 1 = U In words, then, what we have 
shown is that if a is m V, then a -1 must be m U also, or that 
V = U~ l 

11 Let V be a neighborhood of e Then there exists a 
neighborhood of e, U, such that U* C V 

Proof Let V be a neighborhood of e and write 
ee = e 

Now there must exist two neighborhoods of e, and F* 
such that 1 1 Fj C F Let V — F t H Fj Now we have 

U* = UU C Fit * C F 

This result can easily be generalized by induction, to the 
case where if F is any neighborhood of e and n is any positive 
integer there must exist a neighborhood, U, of e with the 
property that {/* C F 



6 . Compactness of Space of Maximal Ideals 


105 


12. Let E be a compact set, 0 an open set, and G a topolog- 
ical group. Suppose E C 0 C G. Then there exists some 
neighborhood V of e such that VE C 0. 

Proof. Let g £ E C 0 and write g = eg. We can find a 
neighborhood of the identity, U„, for each g such that 

U 0 g C O. 

By 11 there exists a neighborhood, which we may assume is 
open, of e, W„, such that IF* C U 0 . Now we can write 

EC U W 0 g 

geE 

as an open covering for E. Since E is compact we must also 
have gi such that 


n 

EC U W 0i gi. 

Consider now the neighborhood of identity 

V = TF 0 I nlF ff! n..- nw, n . 

It is clear that V C for any j. For any element, g, of E 
there must be some j such that 

9 € w aj g, =* Vg C VW 0i g,- C W C C 0; 

whence 


Therefore 


Vg C 0 for any g 6 E. 
VE = U Vg C 0. 

oeE 


13. Suppose Ei and Ei are compact subsets of a topological 
group G. Then the product EiE 2 is compact. 


Proof. Consider the mapping 

G XG-+G 


( git 92 ) 9i92- 



106 


Element* of Abstract Harmonic Analysis 


Since Ei and E> are each compact, the product space E t y E s 
must also be compact Smce the above mapping is continu- 
ous, the image of a compact set must be compact Hence 
E x Et, the image of Ei X E t , is compact 

Exercises 

1 Determine M for the Banach algebra A, of all functions 
analytic m j z j < 1 and continuous in | z | < 1 

2 Suppose 0 is a topology for Af, the set of all maximal ideals 
of the commutative Banach algebra, with identity, X, 
such that A/ is compact with respect to O, and all func- 
tions x (Af) for x £ X are continuous in this topology 
Then prove that 6 coincides with the weakest topology 
on iff for which all x(ilf) are continuous 

3 Let G be a topological group and let g € G, g ^ < Prove 
that there exists a neighborhood V of e such that 

V n gV - 0 

4 Let G be a topological group and let g € G If T is any 
neighborhood of g, show that there exists a neighborhood, 

V of g such that V C V Topological spaces with this 
property are called regular 

References 

1 Nat mark Normed Rings 

2 Pontrjagin Topological Croups 



CHAPTER 7 


The Quotient Group of a 

Topological Group and 

Some Further Topological Notions 


In this chapter we continue with our study of the ele- 
mentary properties of topological groups and discuss certain 
properties of the quotient groups of a topological group. In 
addition some further topological notions are introduced: 
in particular Urysohn’s lemma, the Tietze extension theorem, 
and the One-Point Compaetification Theorem are stated. 

Locally Compact Topological Groups 

Definition. A topological group, G, is said to be locally corn-pad 
if the original topological space, ( G , 6 ) , is a locally compact 
space. 

t 

Definition. The closure of the set of points over which a 
complex-valued function defined on a topological space does 
not equal zero is called the support of the function. 

Theorem 1. Denote by C 0 (G) the class of all continuous, 
real-valued functions on a topological group, G, with com- 
pact support (the function must vanish everywhere outside 
some compact set). If G is a locally compact topological 
group and / £ C 0 (G) , then / is uniformly continuous; i.e., 
for any e > 0 there is some neighborhood of the identity, W, 
such that |/(<?i) — f{gf) I < e for all gig%~ 1 € W. (We note 
that W may always be assumed symmetric) . 

Proof. Let / £ C 0 (G) and let E be a compact set outside 
of which / vanishes. Since G is locally compact there must be 
some neighborhood Fi of e whose closure is compact; i.e., 

e £ Vi and Vi is compact. 

Within any neighborhood of e, however, there must be a 

107 



1 06 Elements of Abstract Harmonic Analysis 

Since E\ and £* are each compact, the product space Ei Y E\ 
must also be compact Since the above mapping is contrnu 
ous, the image of a compact set must be compact Hence 
E\E t , the image of Ei X E t , is compact 

Exercises 

1 Determine M for the Banach algebra, A, of all functions 
analytic in \ z 1 < 1 and continuous in | * | < 1 

2 Suppose 0 is a topology for M, the set of all maximal ideals 
of the commutative Banach algebra with identity, X, 
such that M is compact with respect to 0, and all func- 
tions x(Af) for x € X are continuous m this topology 
Then prove that O coincides with the weakest topology 
on ifr for which all x(M) are continuous 

3 Let 6 be a topological group and let g £ G,gs^e Prove 
that there exists a neighborhood V of e such that 
V n gV = p 

4 Let G be a topological group and let g € G If V is any 
neighborhood of g, show that there exists a neighborhood, 
U of g such that 0 C V Topological spaces with this 
property are called regular 

Refebences 

1 Naimark Normtd Rings 
1 Pontrjagm Topological Groups 



7. Quotient Group of a Topological Group 


109 


which implies f(gg*) — 0, and our claim about inequality 
(1) has been verified. To fit this into the notation suggested 
in the statement of the theorem we let 


gi = 992 for g £ W, 

or, since 

g = gig* 1 , gig * 1 £ w. 

Inequality (1) then states 

\f(gi) - fig*) I < e for all g wp 6 W. (2) 

Since W is a neighborhood of the identity, however, it 
must contain a symmetric neighborhood of identity. This 
neighborhood, since it is a subset of W, will also imply (2) 
and the theorem is proved. 

Remark 1. Note that our definition of uniform continuity 
involved the statement gig^ 1 £ W. In the general topological 
group this is not equivalent to requiring gj'g* £ W. In locally 
compact spaces, however, it is for functions in C 0 ((?). In our 
particular case if we had taken 

U' = {?||/(m) ~f(g 2 ) ! < « for all g* 6 VE) 

we would have arrived at 

I figf) - fig 2) I < « 

for some neighborhood of identity, W, such that g^'gi £ W’. 
It is to be emphasized once again though that local com- 
pactness is the key feature in being able to make this 
statement. 


Subgroups and the Quofienf Groups 
Definition. H is called a subgroup of the topological group, G, if 

(1) H is a subgroup of the group (?; i.e., a, b £ H=> ab' 1 £ 
H, and 

(2) H is a subspace of the topological space G. 

We note immediately that H is itself a topological group. 



no 


Elements of Abstroct Harmonic Anolynj 


Definition A subgroup, H, of the topological group G is said 
to be a dosed subgroup of the topological group if H is a closed 
subset of the topological space, G 
Consider now the coset spacef 

G/R *= {gH\g£G\, 

and consider the natural or canonical mapping 

/ G — » G/H 
g-*gH 

We desire a topology for G/H now that will simultaneously 
make / continuous and make G/H Hausdorff With regard 
to satisfying the first requirement consider the following as 
a possible definition for open sets in G/H 

W C G/H is open if and only if /-‘(W) is open in G If 
the class 

if = {IT C G/H \ is open in G\ 

truly forms a collection of open sets, then certainly / will be 
continuous It will now be shown that IF - does indeed meet 
the required standards 

1 Since / -l (0) = 0 and f~ l (G/H) - G, then certainly 

0 G/H € W 

2 Let | IT a } be a collection of sets in \V and consider 

/-(urn) 

Since 

/-'( u»'.) = u; >( w .) 


t G/H is a group if and only if H is a normal subgroup of the origins/ 
group 



7. Quotient Group of a Topological Group 


in 


and each of the / -1 ( W a ) is open in G, it follows that 
UWa e W. 


3. Lastly, consider TFi, W 2 (L W: 

f-'iWi n W 2 ) = / -1 (W,) (\f- l (W 2 ). 

Since / -1 (Wj) and/ -1 (W 2 ) are each open in G, it follows that 
Wj. n W 2 6 W. 

Thus we are justified in calling W a collection of open sets, 
and it now follows that / is continuous with respect to this 
topology. We will now show that / is also an open mapping 
(takes open sets into open sets) . 

Let 0 be an open set in G. We wish to show /(0) is open 
in G/H ; hence we must show/ -1 (/(0)) is open in G. We first 
note 

f(0) = [gH | g € 0] = OH. 

Suppose now 

9i 6 f-Kf(O)) =>/(<h) 6 }{0) = OH 

which implies there is some g € 0 such that 

f(gx) = gH 
or 

giH = gH 
or 

{g\h | h € H] = {gh | h £ H}. 

Since e 6 H, though, 

ffi 6 \gih\he H\ = {gh\ h£ H). 

Hence there must exist an h £ H such that 

gi = gh 

which implies gi £ OH or 

/- ! (/(0)) C OH. 



112 Elements of Abstract Harmonic Analysis 

Conversely, consider any element of OH , 
gh 6 OH, g € 0, h 6 H 

Since 

/(f*> -iM-gHe; f(0), 

then 

gt 6 /-(/(0)) 

Hence 

/-*(/(0)) = OH 

which is open in G by observation 8 of the previous chapter 
Now we would like to show that G/H is Hausdorff, when- 
ever H is closed Suppose g u gi € G and suppose further 

giH * g,H, (3) 

i e gJI and giH are distinct points of G/H It follows from 
(3) that 

ffi € g%H 

[assuming there is an h, £ H such that gi - gh t leads to an 
immediate contradiction of (3)3 
Since H is closed it follows by observation 8 of Chapter 6, 
that g t H is closed which implies there is some neighborhood 
U of identity such that 

Vgi D g t H ~ 0 

for there must be some open set, V containing g L such that 
V 0 g,H « 0 

(For example, take V — C(gxH) ) Since V is open, IVT 1 
is open and we can take the set U, mentioned above, to be 
Vgi 1 There must exist also a symmetric open neighborhood 
IF C G such that IF* C G We now claim 


WgiH n Wg,H = 0 



7. Quotient Group of a Topological Group 


113 


Suppose not; i.e., suppose there existed w lt w 2 £ W and 
h, hi £ H such that 

Wigih = vhg-ihi => wj'wigi = gJhlq 1 . 


Since W is symmetric, wj 1 £ W ; hence wjwi £ W 2 C U. 
Therefore 

Wi'WiQi £ t/^i 


while 


■gzhhi 1 6 g 2 H, 
which is incompatible with 


Ug n n g 2 H = 0. 


Therefore WgJI n WgJI — 0, which implies 

n Wg 2 = 0 

because, since e £ H, Wg 2 C WgJI and Wg 2 C WgJI. Now 
let 

W = f(Wg i) = WgJI 
W” = /(TFflf,) = WgJI. 

By the above argument W n W" = 0. All we need show now 
is that 

gJS £ W’ and gjl £ W" 
to complete the proof. To this end we note 

f(gO £ f(Wgi) because e £ W 
but 


Hence 

and, similarly, 


/to) = gtH. 

giH £ W' 


gJB £ W". QED 

Before proceeding the following definition is needed. 



1 1 4 Elements of Abjtnjct Hormotvc Aralyiii 

Definition. JST is said to be a normal subgroup of a topological 
group if 

1 N is a normal subgroupf of the group, and 

2 N is a subspace of the topological space 

Consider the coset space now G/N where N is a closed normal 
subgroup of the topological group G We introduce the opera- 
tion of multiplication to G/N by defining 

giNgtN = gtfiN 

and further claim that G/N is a group with this as the law 
of composition This is a standard result from group theory 
and will not be proved here By the preceding statement, for 
any closed subgroup, we know that G/N is a Hausdorff 
space and would now like to show that, in addition the 
operations of multiplication and taking inverses are con- 
tinuous or that G/N is a topological group (Henceforth 
G/N will be referred to as the quotient group or factor group ) 
By observation 3 of the preceding chapter we showed that 
Axioms G3 and G4 were equivalent to ahother condition and 
it is this condition that we shall show G/N satisfies to con- 
clude that G/N is a topological group 
Let IF be a neighborhood of 

QtNpt'N = g&'N 

Hence, with / denoting the canonical map as before, 
gig? € 

We can now assert the existence of neighborhoods, Vi and 
F t , of g t and g t , respectively, such that 

WC /~W =>f(Vi)f{V?) c IF 


t A normal tubgroup N, of a group G baa the defining property that 
gNg~' *> A gN *■ Ng for all <7 As will become evident here the 
normal subgroup in group theory plays a role analogous to that of the 
ideal to rtng theory 



7. Quotient Group of a Topological Group 


115 


Since 

Also, 


9i 6 Vi, giN 6 /(F ,). 


fl? 1 € F?‘=* flfW € /(F^)- 


But, since the canonical map, /, is a homomorphism, we can 
say 

W) = /(F 2 )-» 

or, noting that A must equal A -1 , and Ap 2 = g 2 A, we have 
<7 2 A € /(F,) 

which completes the proof. 


Theorem 2. Let G be a topological group and further suppose 
G to be compact (locally compact). If A is a closed normal 
subgroup, then G/N is compact (locally compact) . 

Proof. First suppose G is compact. Since the mapping 


f: G-* G/N, f(g) = gN, 

is continuous and G is assumed compact, we have written 
G/N as the continuous image (the mapping is clearly onto) 
of a compact set and we can assert that G/N is compact. 

Suppose now that G is locally compact. If so there is some 
open neighborhood of identity, 0, such that the closure of 0 
is compact. The image of e, /(c), is just A and thus 

A = /(e) € f(O) CRO). 

But, since / is continuous, /(O) is comp act while f(0) is 
open because / is open; moreover, /(O) C /( 0) because 
/(0) is a compact subset of a Hausdorff space and, there- 
fore, closed. Finally /(O) is a closed subset of a compact 
space and therefore compact. 



116 


Elements of Abstract Homoroc Anoiyjis 


Directed Sets and Generalized Sequences 
Definition. Suppose S is a partially ordered set with respect 
to < We say S is dtrecied upward if for any x,ymS there 
exists a 2 € S such that x < z and y < z We also write 
z > z and 2 > y To paraphrase this one might say that 
every finite set has an upper bound Similarly one defines a 
set directed downward 

Consider any mapping from a directed set S into an 
arbitrary set X f S — * X / is said to be a generalized 
sequence in X Suppose that X is a topological space \\ e will 
say that the generalized sequence, /, comerges to x € X if for 
any neighborhood, V, of x there exists a t € S such that 
/(a) € V for all s > t, and will denote this situation by 
the following notation 

hm/(a) = x 
s 

In a general topological space, x 6 £ does not imply 
that there is some sequence of elements in E converging to 
x This will be true, however, in any metric space, or, more 
generally, m any space that satisfies the first axiom of 
count&biht} In any case there wilt be some generalized 
sequence of elements in E converging to x Having intro- 
duced the notion of a generalized sequence we would now 
like to introduce a generalized Cauchy sequence 
Definition. Suppose S is a directed set and X is a metric 
space and consider / S — » X / is called a generalized Cauchy 
sequence if for any e > 0 there exists a to £ S suc ^ 

/(t)) < * for all s, t > to where d is the metric on X 
Theorem 3 If / is a generalized Cauchy sequence in a com- 
plete metric space, then / converges, where/ S —* X, S is a 
directed set and A' is a complete metric space 

Proof There exists an s« € S such that for h, h ^ 
t/(fi)) < l/ n where d is the metric on X and n is a 
positive integer Let 

r« » sup(si, s t , s,} 



7. Quotient Group of a Topological Group 


117 


We now claim that {/(r n )j is a Cauchy sequence in X. 
To prove this consider 

dU(r n ),f(r m )) 

and suppose n > m. This implies 

r„ > s m and r m > s m 

which implies 

d(/(r„), f(r m )) < 1/m 

or that { f(r n ) } is a Cauchy sequence. Since X is complete 
there must exist am£ X such that 

lim/(r n ) = x. 

n-+ co 

Suppose now that e > 0 is given and we then choose an 
N such that 2/N < e and such that 

d(f(r N ), x) < e/2. 

Since s > r N implies s > r N > sn we have 

d(/(s),x) < d(f(s),f(r N )) +d(/(r w ),x) 



6 . e 

< 2 + 2 "' 


or 


lim/(s) = x. 
s 


Further Topological Notions 

We first wish to introduce a type of separation not yet 
mentioned. 

Definition. A topological space is said to be normal if it 
satisfies Axiom T<, where Axiom T 4 is as stated below. 



116 Elements of Abstract Harmon c Analys s 

Axiom T< If Jl and K are disjoint closed sets then there 
must exist disjoint open sets one containing H and the 
other containing K 

Some theorems that we will have occasion to make use of 
will now be stated and the reader is referred to any book on 
topology for the proofs 

Theorem 4 Let X be a topological space and let F and 0 
be closed and ojien subsets of X with the property that F C 
O Then A is normal if and only if there exists an open set 
V such that F Q U Q U C 0 

Theorem 5 ( Urysokns lemma) Let H and K be disjoint 
closed subsets of a topological space A r Then X is a normal 
space if and only if there exists a real valued continuous 
function f such that 

1 0 </< 1 and 

2 /(*) =* 0 on H and f(x) * 1 on K 

Theorem 6 (Extension theorem) Let F be a closed subset 
of a topological space X and let / be a real valued continuous 
function defined on F Then A is normal if and only if / 
can be extended to a continuous function on X 

Theorem 7 ( One point campactifica tt on ) Let X be a topo- 
logical space Then there exists a compact space X* con 
taming X and just one other point such that the relatne 
topology on X induced by X* coincides with the original 
topology 

Although we will not prove this theorem here w e will list 
the essential construction involved in the proof and some 
facts about it 

Let j* be anj element not in A (eg lfXisa collection 
of numbers let x* be the color blue) and consider 
X* = X U (x*\ 
and 

Tl the class of open «ets II in A such that <711 is compact 



7, Quotient Group of a Topological Group 


119 


Define 0* to be an open set in X* if 

x* ( 0* (=>0* C2), (a) 

then 0* must be open in X or if 

a:* G 0*, then 0* n X G W. (b) 

Under this construction we now contend that the following 
statements are true (the reader is asked to prove their 
validity in exercise 6). 

(1) { 0*} is a topology for X*. 

(2) X* is compact. 

(3) The induced topology on X agrees with the original 
topology. 

(4) If X is locally compact and Hausdorff, then X* is ’ 
Hausdorff. 

Using these results we can now prove the following theorem. 

Theorem 8. Let £ be a compact set and let U be an open 
set such that E C U in a locally compact, Hausdorff space, 

X. Then there exists a continuous real-valued function, /, 
defined on X , such that 0 < / < 1 and such that f(E) = 1 
and f(CU) = 0. 

Proof. Consider the one point compactification of X, X*, 
and define 

Fi — X* — U, F 2 = E C U. 

We note that Fi fl F 2 = 0. 

It is now desired to show that Fi and F 2 are closed subsets 
of X* under the convention established in the preceding 
theorem and to do this we shall show their complements are 
open. 

Since X* — Fi = U and U is open in X, it follows that F\ 
is closed. 

Since x* G X* — F 2 we must consider X D (X* — F 2 ) 
and apply condition (b) of the preceding theorem: 

Xn (X* - Ft) = X - F 2 = X - E. 



120 


Element* of Abitroet Harmonfc Analyst* 


Now, since X — (X — E) == E is compact m X, it follows 
that Ft is closed in X* By part (4) of the preceding theorem, 
we have that X* is a compact, Hausdorff space and by exer- 
cise 8 we have that any compact Hausdorff space is normal 
Hence, by Urysohn’s lemma, there must exist a real valued, 
continuous function, /, such that 0 < / < 1 and such that 
f(F i) - 0 and /(ft) = I Making the required substitutions 
for Fi and F } and noting that /(X* — U) = {0| implies 
that /(X — £7) = {0} completes the proof of the theorem 

Theorem 9. Let X be a normal space and let F be a closed 
subset of X Let {0,|, t =» 1, 2, • • ■, n, be an open covering 
of F Then there exists an open covering of F, {(/,), i «= 1, 
2, • • •, n, such that V» C 0„ t =■ 1, 2, • • *, n 

Proof Let 

A, - CF U^UO.) 

Then 

ca , ■= f n (nco,) 

which implies 

CAx C O t or CO t C Ai 

Since X is normal there must exist an open set, V, such 
that (see Theorem 4) 

cacvcfc 4, 

Let Vi = CV We now claim that 

CA, Ot, 0, 

covers F, and £?* C Ox To this end consider 
Ux - C? C CV Vi C CV » CV (CV is closed) 
But CV C Oi, hence Ox C 0* 



7. Quotient Group of a Topological Group 


121 


Now, since Ai U CA r = X and 


CUi = V C At =*• CAt C Vi 

we have 


which implies 
and, finally, 


Ai U lh = X 
(At OF) UUtD F 



u Ut D F. 


Proceeding in the same manner we can construct the other 
open sets, t/ 2 , ■ ■ • , U n , having the required properties to 
complete the proof. 


Definition. Let (/,-}, i = 1, • • n, be a family of nonnega- 
'tive, continuous functions defined over a topological space, 
X, such that = 1» for all x £ X. Such a family is 

called a continuous ■partition ( decomposition ) of the identity. 


Theorem 10 . Let {0,-} be a finite open covering of a normal 
space, X. Then there exists a continuous partition of the 
identity, {/,}, such that /,( CO, ■) = 0. 


Proof. By Theorem 9 there exists an open covering, {£/<), 
of X such that (7,- C 0,- for each i. O, and CO,- now con- 
stitute disjoint, closed subsets of a normal space and Ury- 
sohn’s lemma may be applied to assert the existence of 
continuous functions, {</,}, such that 0 < g, < 1 and 


and 


g;(x) = 1 on £7, 
gAx) = 0 on COi. 


It is clear that, for every x, 


J^gAx) > 0 , 



122 Elements of Abstract Harmonic Analysis 

and therefore sensible to consider the functions 


/.(*) ~ 


g.(*) 

2"-ig.(ar) 


which satisfy the conditions 


£/.(*> - 1 


(a) 


/,(*) - 0 if i € CO, (b) 

and, thus, the theorem 


Corollary. Let F be a closed subset of a normal space A 
and let |0,| be a finite open covering of F Then there exist 
continuous functions | /,} defined on X such that 


/.(*) > 0 

£/.<*) = 1 i € f 

»-i 

and 

/,( i) =0 if x d CO, 


Proof It is clear that \0,} and CF constitute an open 
covering of AT By the theorem then there exist nonnegatn e, 
continuous functions 

{/.I, g 

such that 

/,( i) *= 0 on CO , 

and 

g(x) — 0 on F — C(CF ) 

Further, 


£ (Jx) + Q(x) = 1 for any 


x € X, 



7. Quotient Group of a Topological Group 


123 


hence, for x £ F, 


33/<(s) = 1 - 

i=l 


Exercises 

1. Let Gi and G 3 be two topological groups, and f:G l — > G 2 . 
/ is called an iscnnorphism if / is an isomorphism of the 
abstract group Gi onto G 2 and a homeomorphism of the 
topological space G 2 onto G 2 . Give an example of two 
topological groups which are isomorphic as abstract 
groups but not as topological groups. 

2. Let (?i and G 2 be two topological groups, and / : G\ G 2 , 
a homomorphism of the abstract group Gi into G 2 . Show 
that f is continuous or open if and only if it is continuous 
or open at e, the identity of Gi. 

3. A sequence of points { x n } of a topological space X is 
said to converge to the point x £ X (x n — > x) if any open 
set 0 containing x contains all x n for n sufficiently large. 
Let E (Z X. x £ X is called a limit point of E if every 
open set containing x contains a point of E different 
from x. Give an example of a topological space X and 
subset E such that a: is a limit point of E, but there 
exists no sequence in E which converges to x. Prove, 
however, that if X satisfies the first axiom of countability 
and x is a limit point of E, then there exists a sequence 
{x n \ of points of E which converge to x. 

4. Let E C. X, where X is a topological space. Prove that 
x € E if and only if some generalized sequence in E con- 
verges to x. 

5. Let /: S — > X be a generalized sequence in a topological 
space X. x 6 X is called a cluster point of / if for any 
open set 0 containing x and so £ S, there is an s > s 0 
such that/(s) £ 0. Prove that a topological space X is 



1 24 Elements of Abstract Harmon c Analysis 

compact if and only if every generalized sequence in X 
has a cluster point 

6 Justify the contentions in the one-point compactification 

7 Show that if X and its one-point compactification are 
both H&usdorff spaces then X is locally compact 

8 Prove that a compact Hausdorff space is normal 

9 Prove that every metric space is normal 

10 Let X be a locally compact Hausdorff space Show that 
the smallest countably additive class generated by all 
closed sets is the same as the a ring (see footnote on 
p 125) generated by all compact sets provided the latter 
contains X 


References 


1 Pontrjagin Topological Groups 

2 Bourbaki Topologie Generate 



CHAPTER 8 


Right Haar Measures and the 
Haar Covering Function 


In this chapter after drawing on some general results from 
measure theory, the notions of a right Haar measure and a 
Haar covering function are introduced. These concepts will 
then be applied to locally compact topological groups and, 
in particular, it is demonstrated that consideration of Haar 
covering functions is nonvacuous, in this framework. 

Notation and Some Measure Theoretic Resultsf 

The following symbolism will be used in the ensuing 
discussion. 

X: Locally compact Hausdorff space. 

C: All compact subsets of X. 

S: The smallest <r-ring that contains C. Analytically, this 
is the intersection of all v-rings that contain C and is 
the same as the cr-ring generated by C. This collection 
of sets, S, will be referred to as the Borel sets. We shall 
assume throughout that X, itself, is a Borel set. 

t A collection of nonempty subsets of an arbitrary set, T, having the 
property that differences, and finite unions of members of the collection 
are also members of this collection is called a ring of sets. If, in addition, 
T itself is also a member of this class the collection is called an algebra 
of sets. (Note that in an algebra of sets the operation of taking comple- 
ments is closed.) 

If any countable union (as opposed to just finite unions) is also a 
member of the class, then the class is said to be countably additive or to 
be a a-algebra if T belongs to the class, or a a-ring if T does not belong 
to the class. 

A measure is a mapping from a ring of sets into the nonnegative, ex- 
tended, real numbers having the property that the image of the null 
set under this mapping is 0 (this is to rule out the possibility of defining 
the measure of every set to be infinite) and the measure (image) of any 
countable, disjoint union that happens to be a member of the ring is 
the sum of the measures of the individual sets. 

125 



126 BemwrtJ of Abstract Harmonic Analysis 

Definition 1. A mapping ft is called a Borel measure on S if 
M is a measure on S and the measure of any compact set is 
finite 

Definition 2. A right Hoar measure is a Borel measure on a 
locally compact topological group, G, such that 

(1) the measure of any nonempty open set is positive, and 

(2) - „ (E), E 6 S, g € G 

A left Haar Measure is a Borel measure on a locally com- 
pact topological group, G, satisfying (1) above and such that 

tt(gE) *= ft(E) 

Remark I We note immediately that any statement we 
can make about right Haar measures implies the existence 
of an analogous statement for left Haar measures and, for 
this reason, we will restrict our discussion main!y v to right 
Haar measures. 

Remark 2 Property (2) of Definition 2 is equivalent to 
saying that a right Haar measure is invariant with respect 
to a nght translation As an illustration of this, consider the 
additive, locally compact, topological group offered by the 
real numbers, with respect to Lebesgue measure ft is clear 
that in this framework the Lebesgue measure of any non- 
empty open set is positive and, further, that the I ebesgue 
measure of any set on the real line is invariant with respect 
to a right (or left) translation Hence Lebesgue measure is a 
right Haar measure here 

Remark 3 Property (1) of Definition 2 is equivalent to 
specifying that ft is not identically zero 

Proof Suppose there was some open set, 0^0, such that 
*i(0) = 0 Let and consider Og~ x It is clear that Of 1 
is an open set containing e Let V — Og~‘ By property (2) 


ji(0<r l ) = 0 also 



8. Right Haar Measures and Haar Covering Function 


127 


Now let E be a compact set and let h £ E. Then, since e £ 
V, h € Vh and we can write 

E C U Vh. 

heE 

Since E is compact we can write 

EC U Vk { 

«-i 

and 


n(E) < n 



< 'EniVht) = nju(F) = 0 
1 


or 

n(E) = 0. 

Therefore /x = 0 on all of C and it follows from general 
measure theoretic considerations that m vanishes on the er- 
ring generated by C which is S. 

Remark 4. If a right Haar measure is defined, then a left 
Haar measure is (implicitly) defined too. 

Proof. Suppose n is a right Haar measure. Define 

v{E) = n(E-'), E 6 S. 

(Note that E £ S=$ E~* € S.) Now 

v{gE) = niE-'r 1 ) = *(E~ l ) = v{E). 

Hence v is a left Haar measure, it being clear that property 
(1) is satisfied. 

Before proceeding the following definitions and theorem 
will be necessary. 

Definition 3. Let X and Y be arbitrary sets and let & and 
Sy be cr-algebras in X and Y, respectively. A mapping 

T: X-+Y 



128 


El emeriti of Abstract Harmonic Analysis 


is said to be a measurable transformation il, for any set F £ Sr, 
T~ l (F ) € Sx Further, if n is a measure on Sx then we will 
denote 

ti(T~ l (F)) by nT-'(F) 

This is a measure on Sr, called the induced measure (induced 
by T) 

Definition 4. Under the same assumptions about X and Sjr 
a mapping 

f *->/?u(±«| 

is said to be a measurable function f if f~ l (a, «) 6 Sr for any 

a € R U {±<« ) 

Using the same notation as above we can now state the 
following change of variable formula 

Theorem Let X — * V-*iRuJ±«} where T is a measurable 
transformation / is a measurable function, ft is a measure on 
Sx, and nT'~ > is the induced measure on Sr Then, if F € Sr, 

/(T(x»d M (*) - (fiv) dft(T-Hy)) 

in J r 

or 

fTd M *= / fd M T -» 

j t <F) J r 

With this theorem m mind consider 

G^G^RV{± «} 

T g —* gh 

where x is a Borel measurable function and T~ l is given by 
T~ l g -* gh- 1 

t Denoting the smallest * algebra containing all eloecd {open) sets 
of the real line as the Borel tele of the real line we could restate Definition 
4 a-9 follows f tea meaeurable function if the inverse image of any Bore 
aet in the extended real numbers is in Sj 



8. Right Haar Measures and Haar Covering Function 


129 


the o - algebra in G is S, and g. is a right Haar measure. We 
now have 


/ xW dn (g) = j x{g ) dnighr 

J T (G)-0 J n 


but the last term on the right reduces to 


/ x(g) dfi{g) 

J G 

by the right invariance of fi. Thus the right invariant Haar 
measure has led to a right invariant integral. We will now 
show the converse is also true ; namely, that a right invariant 
integral leads to a right invariant measure. 

Denote by ks(g) the characteristic function of E 6 S, 
and suppose 


[ heigh) dfi(g) = f k E (g ) dix(g) = 

J n J n 


But 

Therefore 


gh £ E =» g € Eh~K 
p(Ehr') = g(E). 


The Haar Covering Function 

Throughout the rest of the discussion the following nota- 
tion will be used. 

G\ Locally compact topological group. 

C 0 (G ) : The class of real-valued, continuous functions over 
G with compact support. 

CJ(6): The class of all nonnegative continuous functions 
over G with compact support. 

ip: Whenever this symbol appears, even if with a sub- 

script, it will denote a function from Cf{G) that 
is not identically zero. 



1 30 Elements of Abstract Harmonic Analysis 

With these notations m mind we can now proceed to the 
Hoar covering function 

Consider/ 6 C£(G) and consider all possible finite collec- 
tions of elements 

0u 02, ■•*,{?. e G 
and nonnegative constants 

C$, Ci, C, 

satisfying 

f(0) ^ Ec,p(gg t ) for all g € G (a) 

Assuming some collection exists that satisfies (a) we can 
now define the Haar covering function for f wth respect to as 

(/?)■= inf £>., 

i 

the inf being taken over all sets of nonnegative constants 
associated with some collection of elements that satisfy (a) 
ft wifi now be shown that such elements and associated con- 
stants do exist 

In any case, since v is not identically zero, there must be 
some clement h € G such that v’(h) >0 Since this is so 
there must exist a positive number, «, such that 

<p(h) > t > 0 

Let V — ji? | <p{g) > t) =» ») We can now note two 

facts about V 

(1) Since it is the continuous inverse image of an open 
set it must be open, and 

(2) U 9 * 0 because h E U 

Since G is locally compact there must be some neighborhood 
of A, IT, such that If’ is compact Consider the neighborhood 
of h defined by V = V n If Now, since f' C If > it follows 



8. Right Haar Measures and Haar Covering Function 


131 


that V is compact. But 

VC uc { g\ v (g ) > ej 

SO 

VC {g ! <p(g) > e}. 

Hence 

inf <p(g) > e > 0. 

oeV 

Since V C V we also have 

m 9 = inf <p(g) > inf <p(g) > 0. 

gtV 0CV 

Now, since there must exist some compact set, E, outside of 
which / vanishes and a continuous function must actually 
attain its extrema on a compact set, there must exist a posi- 
tive number, M f , such that 

fig) < Mf, all g e G. 

For any g € E we can write 

g € Vh-'g 

where h is an arbitrary element of V. Letting g range through 
E, it is clear that the sets { Vlr'g] cover E. Since E is com- 
pact there must exist gt <E G such that 

n 

EC U VgJ 1 . 

1=1 

We now make the following : 

Contention. For every g € G 

" Mr 

fig) < JL — viggi)- O) 

«-l m r 

To prove this we must consider two possibilities: 

(1) If g 6 CE, then f(g) = 0 and (1) is satisfied. 



132 


Element* of Abstract Harmonic Analysis 


(2) If g € B, since the { l g7' I coscr B, there must exist 
an t such that 

g € T'g7 ! or gg, € V 

which implies 

vigg.) t 

Adding positive terms to the left hand side, 


5Zv>(W.) ^ n * 

which implies 

£ WOTO 5, J 

1-1 w * 

Therefore 

M,£tlnA > m, >/((,) 

.-I 

which completes the pnJof 
Remark 5 This proof immediately implies 


(/ *) < 


fiM/ 

m. 


In the following theorem we list and prove some rather 
immediate consequences of the definition of the Haar co\ cr- 
ing function 


Theorem 1. If / £ CJ((?) then the following is true 

(1) Define f k (g) ** f(gk) Then (/* <i) - (/ <f>) 

(2) If a > 0 then (af #>) = «(/ 

(3) </, +/, ?) < (/, *) + (/, 

(4) (/ ?j) < (/ 

Proof (1) Let g t and c, be such that (a) is satisfied or 

/W < 2>Wsf.) <«) 

( 



8. Right Haar Measures and Haar Covering Function 


133 


and consider 

Mg) = f(gh) < Xav (<?%)• 


Thus the elements c it hgi satisfy (a) for f h and we have 
|c» l/(g) < Xc<¥>(09i), for all gj 


C 


di\Mg) < ^Ldi<p(gg<), for all g . (2) 


We wish to show this also goes the other way around and 
for this purpose consider jd,}, jg,-J such that 

Mg) < Udiviggi) => f(gh) < J^di<p(ggi). 

i i 

Letting y = gh we have 

f(y) < Jldi<p(yfr 1 g<) for all y £ G 

i 

to reverse the inclusion in (2). Therefore 
(fh - <p) = ( /'• v ) • 


(2) If 


f(g) < J2ci<p(ggi), 


then 


af(g) < ^acivigg.) , a > 0 


and conversely. Hence 



?34 Elements of Abstract Harmonic Analysis 

(3) For any t > 0 there must exist c„ g, such that for 

/lCfl) < J2c,<p(gg,) and £c, C (ft v) + t 

by the definition of in/ Similarly there must exist d,, h, 
(i =- 1, 2, k ) such that 

t i 

/i(f>) < and £d, < (/, v ) + * 

We non have 

(/i +/»)($> «/.(?) +/.(g) < I>.Kgg.) + 

Call 

di = c*+i Aj = g.+j 

dk =* c*+t A* = g«+* 

Hence 

(/. +/. v») 5! 

= X>. + XX < (/. o) + (/, v) +2« 

But, since c was arbitrary this implies 

( fi + /* «0 < ( h <f) + ( /. v?) 

(4) There exist c„ g, such that 

/(g) < XX^igg.) and 2X <(/**)+« ( 3 ) 

and dk, g* such that 

vi(g) < XXv’i(ggi) and XX < (<n «*.)+* 

‘ * (4) 



8. Right Hoar Measures and Haar Covering Function 


135 


for any e > 0. Rewriting (4) we obtain 
vi (gg<) < 

k 

Substituting this in (3) yields 

f(g) < H'''iY J dk‘Pz(gg>gk ) . 

i k 

Hence 

(/:«*) < Halid, 

t k 

and 

(f-v 2 ) < 1 ) 4 - e) ( ( vi ■ Vi) + «) 

and, since e was arbitrary, 

(f- <pt) ^ ( f- <Pi) ( vd • 

Theorem 2. If / G C%{G) and / ^ 0, then (/: <p) > 0. 
Proof. Consider c; and </; such that 

fig) '< H c ‘<p(<J0i), for all g £ G. 

i 

Certainly we also have 

fig) < sup <p(g)H c <> for any g, 

qcG i 

which implies 

sup/(ff) < sup <p{g) H c ‘- 

gcG oeG i 

Therefore 

sup / < 

sup (p ~ 

Thus for any associated sum, H< c <> SU P // su P V a l° wer 
bound. Therefore 



0 < < (/; p). QED 

sup <p 



1 36 Elements of Abstract Harmonic Anclyiii 


Consider now some function/# £ C£(G) such that/# & 0 
We define 


l.U) ■ 


and note the following properties which all stem directly 
from Theorems 1 and 2 


Remark 6 If / 5*? 0, then /,( /) > 0 

(A ») = (/ y) 

(/. v) = (A d 
If a > 0, then 7,{c/) = a/,( /) 


Remark 7 /*(/») *= 
Remark 


- M/) 


Remark 9 l/(/# /)</,(/)<(/ /.) 

To prove this we note that part (4) of Theorem 1 implies 


(/ v) 
(A #>) 


< (/ A) 


and also that 


(/ ») > _L_ 
(/• e) ~ (A /) 


QED 


Let U be some neighborhood We will denote those func- 
tions in CJ(G) that vanish on C U by Fv Fv is not empty 
To see this apply Theorem 8 of Chapter 7 with E = |p} 
where V is a neighborhood of g 


Theorem 3. Let t > 0 be given and Jet/i /j •*•,/« 6 C} (G) 
with the property that £•-» /,(p) < 1 for all g € G 
Then there exists 17, a neighborhood of identity, such 
that for all <p € Fv and any / 6 C# (G) 


tu/A) </.(/)(!+«) 

Proof By Theorem 1 of Chapter 7, there exist neighbor- 
hoods of e, U„ such that for i = 1, 2, • • n 

i/.w - /.<«') I < «/" 



8. Right Haor Measures and Haar Covering Function 


137 


for g'g~ l € £/,-• Thus for V = we have 

I Mg) - Mg') I < - , g'g~ l € V 

n 

for every i. By the existence statement of this chapter there 
exist cu and g k such that 

Kg) < Z ]ck<p(ggk) (5) 

k 

and 

Z c * ^ (/ : v) + v, v > 0. 

k 

Since <p 6 Fv then certainly <p(ggk) — 0 if gg k $ U. Hence 
in (5) nothing is lost by disregarding all those k for which 
gg t $ U. Denoting by 

z* 

k 

the summation in which only those k such that gg k 6 U we 
can rewrite (5) as 

Kg) < Z *Ckv>(ggk). 

k 

We now have 

Mg) Kg) < H*Ck<p(gg k )fi(g ) for gg k 6 u and any i. 


( 6 ) 


However, gg k £ U is equivalent to 

gig?)- 1 € u 

which implies 


\Mg) - Mis 1 ) I < 

Using this we can replace (6) by 

Mg)f(g) < T,*ck(fi(g T 1 ) + 



13B Elements of Abstract Harmonic Analysis 

Since adding a few zeros certainly cannot hurt w e can w nte 
/.(»>/(?) ^ for e\ ery g 6 0 

Therefore 

(/./ «0 < £&(«£') + ‘A 

which implies 

£(/./ rt < £ £a(/.(«r‘) + 0 

But, since for any g £,/<(g) < live can say 

£(/./ *) < &(1 + «) 

In addition however, we had 

£c* <(/«»)+•', 


hence dividing through by ( /« <p) (ft ^ 0) gives 


T (fJ v>) < 

. (/. *>) ~ 


(/ 1?) + V 
(/» *) 


(l+«) 


w here v is arbitrary Letting v go to zero yields 


£l,IM </.(/)(! +«) 


which is the desired result 

It is exceedingly important for the later development to 
note that one cannot Jet t become arbitrarily small while 
stilt retaining the same v>, * determined If which, m turn, 
affects ip It is granted that for any t there is some U and, 
consequent!} , •some <p that will work but the U and <p that 
work for e t may well not work for an t 3 < «i 



8. Right Haar Measures and Haar Covering Function 


139 


Although the next two theorems we wish to present are 
rather top-heavy with hypotheses we beg the reader’s in- 
dulgence for these theorems will play a vital role later on. 

Theorem 4. 

The Conditions. 

1. G locally compact topological group. 

2. h,h, •••,/« € C 0 + (C). 

3. v, X > 0. 

The Conclusions. There exists some neighborhood, U, of 
identity such that 

for all <f> € Fv and all X, £ [0, X]. 

In words: I v of the linear combination is less than or equal 
to the linear combination of the I v which, in turn, is less 
than or equal to I v of the linear combination plus some 
arbitrary positive number. 

Proof. The left-hand inequality follows in a straightfor- 
ward manner from part (3) of Theorem 1 and Remark 8. By 
the second condition, for each /,-, there exists a compact set, 
Ei, such that /< vanishes on CEi. Clearly then all of the /,• 
vanish on 



We further note that E = U"_i2?i is compact. Let h be a 
function in CJ (G) which has a positive minimum on E. 
Before proceeding we will prove that such an h exists. 

Let Y be an open set containing E. Since G is a locally 
compact Hausdorff space there must (see Halmos 0?]) exist 
open and compact sets, Uo and Co, such that 

E C U 0 C Co C V. 



140 


Elements of Abstract Harmonic Analysis 


Also, in a locally compact Hausdorff space, in a situation 
like this (see Theorem 8 of Chapter 7) there exists a con- 
tinuous function, h, such that 

h(E ) = 1 and h(CU 6 ) = 0 

•which is a suitable function 
Picking up where we left off, let 

/ - £»./. + <h 


where <> Oand 


Kf,/f on E 

0 on CE 


where it is noted that, by the choice of h,fc an never be zero 
on E We will row prove the following 


Contention. g t f — X,/, everywhere 

1 This is clearly true on E 

2 Since both g x and /, are identically zero on CE the 
contention is proved 

Since 


32xi. 

X«/. + th 




we can apply Theorem 3 to assert the existence of a neigh- 
borhood of identity, U, such that for all tp € Fv 


1 - 32i.0-.f-) - £/,(/».) </.(/)(!+') 

i i i 


+ <!■)(■ +«) 



8. Right Haar Measures and Haar Covering Function 


141 


By the first part of this theorem we also have 

£l*(X,/,) < (l^IX/i) + el v my 1 + e) 

and we claim that e can be chosen small enough so as to make 

+ <?l v {h) < V 

which will then yield 

IX(x,<7,) < + *■ 

Some care must be taken in this argument though, because 
(p depends on e, i.e. for our argument to be valid we must 
show that 


I v (h ) and I v 



remain bounded as e goes to zero. We first note that 
I v (h) < (h:fo) < » 

for any <p, by Remark 9 following Theorem 2. 

Finally, we can write 


< 2>.(/.:/o) < TMU-fo) < «» 

' i / i i t 

which shows that /„( 2>X./.) is bounded for any <p and 
completes the proof of the theorem. 


Theorem 5. 

The Conditions. 

1. G a locally compact topological group and / 6 C£(G) 

(/ ^ 0 ). 

2. U a neighborhood of e such that |/(fir) — f(§) I < e 
(e > 0) for g 6 Ug. 



1 42 Elements of Abstract Harmonic Anatys s 

The Condos ons For any 8 > « and any $ € Fv ^ 0) 
there exist 

Si a ff»6G 

and positive numbers 

Cj o c, 

such that 

- £ciKW<) j < 5 for all g € G 

In words this says that in some sense we can approxi 
mate f(g) 

Proof Vi e claim that for all g g 

( /(a) - <)Hw ') <f(a)Uaa *) < (/(?) ') 

(1) 

To prove it we consider two cases 

1 gg 1 6 U In this case the validity of the above in 
equality is obtained directty from the hypothesis 

2 gg 1 £ U Since gg 1 $ U => $(gg ') ■* 0 the inequality 
is again satisfied hence we have it for all g g 

\\ e now choose a » > 0 such that 

(/ '('*)•’ < 8 — t (2) 

where \fr*(g) — iHp -1 ) and assert the existence of a sjm 
metnc neighborhood of the identity 1 C G such that 
for g~‘g € T 

I Ms) - Ms) I < - P> 

and such that 1C"" here 11 is compact 
The last statement is valid for the follow mg reasons Since 
it 6 CJ(C) it must be uniformly continuous (Theorem 1 
Chapter 7) i e for any v > 0 there exists a neighborhood of 
identity 1 ( such that 

I Ha) - Ha) I < *■ for <r x g £ J ? ( 4) 

Take 1 1 f) U = I j Since G is locally compact there is some 



8. Right Hoar Measures and Haar Covering Function 143 
neighborhood of e, F 2 , such that F 2 is compact. Take 

w = Fi n f 2 c f 2 

and note IF C F 2 and therefore IF is a compact neighbor- 
hood of e. But any neighborhood of e contains a symmetric 
neighborhood of e (see observation 10, Chapter 6) ; therefore 
there exists a symmetric neighborhood of e, F, such that (4) 
is satisfied for g~ x g £ F and 

F C IF. 

Since / £ Co'(G) we can say there is some compact set E 
such that 

F = \x\f{x) > 0} C E. 

Since e £ F, then certainly 

ECU Vg. 

CtE 

Since E is compact there must exist g { , i = 1, • • • , n, such 
that 

E C U V 9i . 

i=i 

By the corollary to the last theorem in Chapter 7 we are 
assured of the existence of functions, hi 6 Cq(G), such that 

n 

^2hi = 1 on E 

t=i 

and hi = 0, on C ( Fg.) . (To prove this consider the one- 
point compactification of X, X*, which is a normal space in 
which E is closed because E is compact in X. Now apply 
the corollary to the last theorem in Chapter 7; note hi 
vanishes on C(Fg t ) and therefore on C(IFg,), but TFg, is 
compact so hi £ C„ (G) ) . 

It is now claimed that 

hi(g)f(§) (tigr 1 ) - v) < hi{g)S{g)Hggl l ) 

<hi(g)f(g) OKsir 1 ) + *0 

for all g, g £ G. (5) 



Element of Abifr-ort Harmonic Anolym 


UA 

Suppose g £ CE Since / vanishes on CE it is clear that 
(5) holds in this case 

Now suppose g $ CE or g € E If g e E, then for some 
given t, either g € Vg, or g $ Vg, In the latter case then, 
k,(g) — 0, m the former case, 

g € Vg.=> ggT 1 € V => gg-'QQT 1 € V 

=* I iKefl" 1 ) - Mm 7 1 ) | < v 

by (3) which proves (S) Since (5) is true for each » we have 
also 

ZX($)/(y) (MssF') - *) < ZXfpXftyWw; 1 ) 

^ £M$)/(y) (iKw -1 ) + »>) 

Noting that <p vanishes outside U and \f(g) — f(g) ( < « 
for gg~ x € V, the above statement implies 

&(«[</<»> - >H(lC) - -/«)] 

< 2>.(j)/(5)*(ra7') 

< 2>.«)t(/(p) + «>*(«-> + rf(S)3 (0) 

lf/(p) ~ < > 0, then it is claimed that (0) yields 

((/(g) ~ *)$*(gr l ) - >/($)) 

< ( f(g ) + + »/{?) ^ 

Only the left half of (7) will be proved, the right half 
following in a similar manner Once again the problem is 
broken up into two parts 

1 g € Vg 

By condition 2 of the hypothesis and our assumption that 



8. Righf Haar Measures and Haar Covering Function 


145 


J(g) — t > 0 we have 

m > Kg) - 6 > 0=* g € E=> jtfiiffi = 1. 

»“1 

Substituting this into (6) yields the left-hand side of (7). 

2. g <£ Ug=* <p{gg~ l ) = 0, 

which immediately yields the left side of (7) . 

It is now claimed that for every g we have 

(Kg) - UM*) - vi r U) 

< ij(jjl>*(g^- t )Mg)KS)y j 

< (Kg) + «)/,(**> +vi,(f)\ 

Here too only the left side of inequality (8a) will be proved, 
the right side, (8b), following in an analogous manner. 

Proof of (8a). If /(g) — « > 0. (7) implies 

U(g) - 0 r(gg~ l ) < »f(g) + XXtfW’farOT)- (9) 


(8a) 


(8b) 


But (see exercise 4 of this chapter) 

Mg) <M§) for all g 

implies 

UK) < UK). 

We also know that, in general, 

Uh)=Uf), h feed > 

and 

Uh+K) < UK) + UK- 

Letting g be held fixed, these results when applied to (9) 
yield 

(Kg) - «)!„(**) < »Uf) + i^Jjt*(g<rWS)KS)) 



Elements of Abstract Hannon c Anatyi 


1 46 


which was desired Noting that (8a) is clearly valid if 
fid) t < Q, the result (8a) is proved Nov, , since by part 
(4) of Theorem i, 


and 


ur> 

w> 


(/ y) 
(i* v) 


< u 


*') 


u *•>>■ < s - 


there must exist a {1 < 8 such that 


u *•> - 


Dividing (8a) and (8b) by /.(^*) and using this result gives 

m -« = /<») -<-«-<> s/(?) 


< r (r ¥M2MbM \ 

- '\r ) 


<As) + < + 


/.(/) 

r.t*-) 


< As) +<+«-.- As) + B CIO) 

We now wish to apply Theorem 4 of this chapter and to 
simplify matters will list the correspondence of symbols 
used here with those mentioned in Theorc'i 4 


Theorem 4 Here 

/. 6 CtiG) h,f 

\ sup 

i 


^*(g<r») 
/.<*•) 
s~ e 


X, 



8. Right Haar Measures and Haar Covering Function 


147 


The fact that “X” < “X” here follows from Remark 9. 
Noting that Theorem 4 is applicable, there exists some 
neighborhood of identity, V, such that for all <p 6 Fv, 



IM*) 


< E 

t 


lfr*(g.'g~ 1 ) 

/,(**) 


hVuf) 


Let 


+ S - 0 


IM*) 

for every g. 


(H) 


IM*) Ci 

in (11). Then we can replace (10) by 

m - a < TfifTig#- 1 ) < f(g) + 0 + * ~ P = fit) + 5 

X 

or 

/(fit) - 8 < Ecif(ggri) < f( g ) + s. QED 


Summary of Theorems in Chapter 8 

1. If / 6 Cf(G), then the following is true: 

(1) Define f h {g) = f(gh). Then (/*: to) = (/: to). 

(2) If a > 0, then ( af : <p) — a(f : <p). 

(3) (/i + /s'- to) < (/r- to) + (/* : to). 

( 4 ) < ( f'- <Pi) (<Pi'- <Pz) • 

2. If / 6 Cf(6) and/ ^ 0, then (/:<?) > 0. 

3. Let 6 > 0 be given and let/i, />, •■*,/» € Co" (G) with 

n 

the property that XUi/i(?) 2= 1 for all g € (?• 



148 


Elements of Abstract Harmonic Analyst 


Then there exists, U, a neighborhood of identity, such 
that, for all <p 6 Fv, and any / € Cf(G), 

£/,(//<> < /,(/)(i + <) 

4 The Conditions. 

1 G a locally compact topological group 

2 /«/,•••,/. 6 Cf«?) 

3 »>, X > 0 

The Conclusions There exists some neighborhood, U, of 
identity such that 

r,(l>./.) < < r.( + >• 

for all <fi € Fv and all € £0, ^!1 

5 The Conditions. 

] G a locally compact topological group and / 6 Cf (G) 
(/ * °) 

2 C/, a neighborhood of c such that j /(g) — /( g) I < ‘ 
(« > 0) for ff 6 Ug 

The Conclusion For any 3 > t and any it £ Fy(\t 0) 
there exist 

tfi. gt g* £ G 
and positive numbers 

C| Cl, • , c« 

such that 


\J{g) - T!.ceHgg,) 1 < 6 for all Q € 0 



8. Right Hoar Measures ond Haar Covering Function 


149 


Remarks 5-9 Inclusive 


5. 


(/:*) < 


nMj 

m f 


6. If / ^ 0, then I v ( f) > 0. 


7. /„(/*) 


(/»:*>) = (/: y) 

(M<p) (fo'-v) 


8. If a > 0, then I?(af) = al f (f). 


Exercises 

1. Let js be a Haar measure in G. Show that G is discrete if 
and only if p({gj) ** 0 for some g £ G. 

2. If ft is a Haar measure in G and a is any positive real 
number, show that an is also a Haar measure. 

3. Let n be a Haar measure in G and let E\ and E? be two 
compact subsets of G such that n{E{) — n(E 2 ) = 0. 
Does this imply that n(EiE 2 ) = 0? 

4. Consider /i,/ 2 , <p £ Cq (G ) , and <p ^ 0; prove that/i(g) < 
Mg) for all g € G implies that I v ( fi) < I p ( / 2 ) . 

References 

1. Naimark, N armed Rings. Especially pp. 360-366. 

2. Loomis, An Introduction to Abstract Harmonic Analysis, pp. 113-116. 

3. Halmos, Measure Theory, p. 218, Theorem D. 



CHAPTER 9 


The Existence of a Right Invoriant 
Haar Integra! over any Locally 
Compact Topological Group 


Using mainly the results of Chapter 8 it will be shown 
here that over any locally compact topological group, G, a 
right invariant Haar integral exists Our first theorem will 
be devoted solely to proving this statement The integral 
will be shown to exist only over the class of continuous func- 
tions with compact support Co(G), however, and this 
alone is not quite to our satisfaction What we want is a 
larger class of functions to work over, or, equivalently, a 
larger class of sets to which the right invariant Haar measure 
defined by the above integral may be applied Hence after 
having proven the existence of a right invariant Haar inte- 
gral over Co(G) we wish to extend it to a larger class f 

Two avenues of attack are available to us as possibilities 
for providing the desired extension 

(1) We might concentrate our efforts on only the integral 
and not concern ourselves directly with the underlying 
measure extension that we are brmging about at the same 
time In particular it is pointed out that all the ingredients 
necessary for the Darnell extension approach are there and, 
hence this possibility is at our disposal but is not elaborated 
upon further here In Chapter 10 more on the 1 integral 
approach will be said 

(2) Alternatively we might proceed in a measure theoretic 
fashion and concentrate our efforts primarily on the exten- 


t The desire to do this is more easily understandable wlicn one com 
pares this to the severe limitation of the Rtornan integral of being re- 
st noted to functions that are continuous almost everywhere with re- 
spect to Lebe«gue measure and ita extension to the I^beegue integral 
150 



9. Right Invariant Haar Integral 


151 


sion of the measure, and in this chapter we shall pursue this 
possibility. The integral defined in the existence theorem 
will be used as a springboard for obtaining a measure. An 
appendix listing some pertinent results from measure theory 
is included at the end of this chapter for the reader’s 
convenience. 

We wish now to proceed to our main result of proving the 
existence of a right invariant Haar integral on the linear 
vector latticef of all Borel measurable functions on a 
locally compact topological group, G. First we will prove the 
existence of a right invariant Haar integral over the linear 
vector lattice of all continuous functions with compact 
support over G. 

Theorem. Let G be a locally compact topological group. 
Then there exists a right invariant Haar integral over 

C«(G). 


Proof. We introduce the notation 


{e>}: All those elements in C${G\ which do not vanish 
identically and 

S: ip (E [<p] which do not vanish in some neighborhood 
of the identity, e. 

It is now claimed that S is partially ordered as follows: 
Let <pi, <p 2 , 6 S. Then define 


if 


1 Pl > <P2 


! g\Mg ) = 0} d \g\M = o). 


It is also claimed that <S is a directed set for given any 
<Pi, <p2 € S, then, since neither vanish in some neighborhood 
of e, <p\<p2 € S) and 


Wf! A <pi, 


<pi<P2 2! e>2' 


t See p. 159. 



1 52 Bementt of Abstract Harmonic Analysis 

Let / € C*f (G) and consider the mapping 
S~* R 
v -/,</) 

Thus {/„(/)} is a generalized sequence of real numbers and 
we will show that it is a actually a generalized Cauchy 
sequence It is claimed that (with Fv as on p 136) 

{if, for t > 0, there exists some neighborhood of identity, U, 
such that 

1 /,.(/) - /«(/) 1 < « for all n € Ftr) (•) 
then {/»(/) j is a generalized Cauchy sequence which implies 
lim /,( /) exists 

First it will be shown that condition (•) does actually imply 
that {/,(/)) is a generalized Cauchy sequence and then ne 
will show that such neighborhoods of identity exist To proie 
the former contention consider some neighborhood of the 
identity, U Since we are in a locally compact space we can 
assert the existence of another neighborhood of e, V, such 
that the closure of V is compact,} and V C U Further there 
must exist open and compact sets l/» and C 0 , such that 
V C Vo C C U (see bottom of p 139) and some con- 
tinuous function <p, that assumes the value 1 on f and 0 
on CUt> Assuming (•) to hold and U to be the U mentioned 
in (•) then the inequality mentioned in (•) will hold for 
all <pi, & in Fv Then, certainly, <p € S and 

I /,*(/) -/„(/)!<« 

for tp\, <pt > <p where <pi n € S Thus assuming (•) docs 
imply that I /*(/)) is a generalized Cauchy sequence We 
will now prove that ( •) holds 

If /, fa € Ct(G),f 0 & 0, then there exists some neighbor- 
hood of e, V, with compact closure, such that 

I /(p) — /($) | < <i and \Mo) -/»($) I < ** 

t This follows from the local compactness of O and exercise A of 
Chapter 6 



9. Right Invariant Haar Integral 


153 


for all g 6 Vg where a is an arbitrary positive number. 
Since, in the subsequent discussion, we will be interested in 
“small” values of ex, we will assume now that ex is less than 
one. Now let t/ € Fv be not identically zero. By Theorem 5 
of Chapter 8 there exist c, and g, such that 

w(g) = | /(g) - 2 Zc<t(ggi) | < 2ex. 

i 

where i ranges from 1 to n. 

By Remarks 5 and 9 of Chapter 8 we can say 


and 


I v ( w ) < («>:/o) 


( w: fo) 


„ max to, 

< k < 


m f . 


where k is some integer and m !s is inf / 0 (over some open set) 
which is positive./ We would now like to show that k does 
not depend on ex (for all «i < 1 that is) . Once this is estab- 
lished we will have shown that I v (w) is bounded as e t goes 
to zero. Since k depends only on the compact set outside of 
which w vanishes (see p. 131), if we can exhibit some com- 
pact set outside of which all w{g) vanish, regardless of ex, 
then we will have demonstrated the independence of ei and k. 
To this end consider 

F = \x \ /(*) > 0}. 

Since / £ CJ (G) it is clear that there is some compact set, 
E, such that F C E, and, further, the g, mentioned before 
must be such that 


EC U Vgr> 

by the proof of Theorem 5 of Chapter 8. Hence for each i, 
Fgv 1 n E 0 

t See the original discussion concerning the nonvacuousness of Haar 
covering functions, pp. lpO-132. 



154 


Elements of Abstract Harmonic Analyt s 


which implies 

lg 7 * n E j* 0 

We can now assert the existence of a i 6 V such that 

*87* * x € E=* 177 s - vr'x 6 f-'E 
or 

07 1 e V-'E 

It is now contended that all to(p), for all «i < 1, vanish out- 
side the compact set 

Vi V?E U E 

where V\ is the neighborhood corresponding to «i » I, and 
we will prove this by showing that each of the terms, /(<?) 
and S» C -V'({?P«) is zero outside this set 
Clearly if 

g <£ ViV^'E U E 

then 

gi E 

which implies /(ff) = 0 We would now like to show that a 
sufficient condition for each of the f(gg t ) to vanish is that 
9 $ Yjtft 1 

To this end suppose 

9 § Vi07‘ Vtf? =* f?0. t l r i 

Since jfr 6 Ft-, then y p(gg t ) = 0 if p $ Vi97 l 
In our case we have 

g f \ t V?E, 

but, since 07 ‘ € Tr 1 #, then 

17 i I iS7‘ 


Therefore 


Hggt) - 0 



9. Right Invariant Haar Integral 


155 


which implies w(g ) = 0 outside ViV^E U E, and we have 
established the required independence of k and e t . 

Before proceeding to our next step in the proof we wish 
to note some facts which come from the theorems in 
Chapter 8: 

1. It is clear that there is some compact set, E', outside 
of which both / and f 0 vanish. 

2. In Theorem 5 some functions, hi, were constructed. 
With regard to this construction (with respect to E') it is 
clear that the same hi will work equally well for f and f 0 . 

3. By Theorem 4, there must exist some neighborhood of 
e, U, such that for all <p £ Fu 

h(ll c d{ggi?) < 

= (Zc,)/ v W < I 9 (l ™ (1) 

where we have the following correspondence: 

Theorem 4 Here 

ei/w/c 

h(hl) 
sup * = sup — 

_ yjuj) 

Ci 1M*) 

tiggi) 

4. It is clear that an analogous statement to (1) can be 
made about /o because of the restriction imposed upon V; 
the same U and the same ^ will work for /o, the only change 


v 

X 

X, 

Si 



156 


Elements of Abstract Harmonic Analysis 


necessary being fo replace the c, by 




This done we will now perform some manipulations with 
(1) knowing that, in the end result, we need only replace 
the c t by d, and / by / a to get the same result for /„ By the 
definition of u>(g) we have for any <p € Fa, that 

1,(1) - — l < < 1.(1) + — t 

mj, \ , / tn/, 


Using this and (1) together we have 

i.d) - < (&)/.<« 


< un + — i + — 

m/, m,. 


for v € F v 


Let = (2k + l)ti/fn f . We have 

1 ,( 1 ) - « < ( ) < 1 .( 1 ) + o (2) 

But, as previously noted, we now also have 

1.(1,) - « < (£ijl.(*) < 1.(1.) + ( 3 ) 

Using (2) and (3) together now gives 


o - «> - « < i.u) < g~o + ..) + ■> 

Adding 1 throughout and assuming <* < i we have 

X>. „ /,(/) + i 

zj , + ,s ' f-v 

< 2 (/,(/) + 1) < 2((/7»> + 1) 

for any <p € Fa. 



9. Right Invariant Haar Integral 


157 


Hence if ^2 € F v , 

I !«(/) - /„(/) I < 2^1 + 

= 2(2 k + 1)— (2) ((/:/,) + 1). 

Letting ei — » 0 now yields the desired result; namely that 
{/*(/) } is a generalized Cauchy sequence and, since R is a 
complete metric space, 

lim I v (f) —1(f) exists, 
s 

This done, we will now show that 1(f) is indeed a right 
invariant integral over Cf (G) to complete the proof. 

1. In general if / £ CJ(G) and / N 0, then 

'• (/) a zb) > 0 

for all <p. This implies that 

1(f) >0 if / > 0 and /NO. 

2. Since, if h is fixed, 

I'UW) = I,(f(g )) 
for all <p we have that 

I(f(gh)) = I(f(g)) as well. 

3. If ci, Oi > 0, and v > 0, there exists some neighborhood 
of e, U, such that for all <p £ FV 

I f«»(ci/i + C 2 / 2 ) — cil r (fi) — cfl v (fi) | < r 

by Theorem 4 of the last chapter. Thus 

1 f( c i/i + C 2 / 2 ) — cil(fi) — c*l(h) ) < v 

where v is arbitrary. Therefore 

/(Cl/l + C 2 / 2 ) = cj(fi) + (hi (ft)- 



158 


Element* of Abstract Harmonic Analysis 


Finally if we take 

J + (g) = 

and 

/ Iff) - 

and then take 


m 

0 

-Jiff) 

0 


if f(g) > o 

otherwise 
if f(g) < o 

otherwise 


/(/) -/</+)-/</) 


we see that 1(f) ail] hate the properties mentioned aboie 
over all of C 9 (G ) This completes the proof 

Haling constructed the aboie integral oter C»(G) our 
de«ire now is to extend it and as pointed out earlier manj 
approaches present themselves 

First we will bneflj show that all the ingredients nccessarj 
for the Darnell extension approach (as used by Loomis [1] 
in Chapter 3 of his book for example) to extend the integral 
from just C 9 (G) to the Batre classes are present 


The Darnell Exlenuon Approach 

Noting that Co(G) is a real vector space we are justified 
m calling the I ( /) defined above a linear functional Further 
since 1( f) has the additional propertj that 

/ > 0 =>/(/) > 0 

we will call I a posihie linear functional 
Consider now a sequence of functions fi /« of 
C«(C) that contergcs monotomcaH> decreasing!} to zero f 
Since 

/.</.) < </. /.> < — ^ 

m, 


t It is clear by D n « theorem that s nee the /. are coot nuotis and 
have compact support that they must converge un forroly 



9. Right Invariant Haar Integral 


159 


where k is some integer, which can be chosen independently 
of n, it is clear that I v ( /„) converges monotonically to zero 
which implies that /(/„) converges monotonically to zero 
too. Hence 


monot monot 

fn >0 =»/(/„) * 0 

and we shall refer to this as property (M). We define now 

/A 0 = inf(/, g) 
fVg = sup (/, g) 

where/, g £ C 0 (G). It is clear that the functions / A g and 
/ V g are each members of Co(G). Denoting the above 
operations as the lattice operations we can paraphrase our 
last result by saying the Co(G) is closed under the lattice 
operations or that C a (G) is a linear vector lattice. We can now 
summarize these observations about C 0 (G) : 

1. C 0 (<?) is a linear vector lattice, and 

2. 7 is a positive linear functional on C 0 (G) with property 
(M). 

These facts will now allow us to define the Daniell exten- 
sion of 7, I e , to the Baire classes. Further since, denoting 
the identity function of Ca(G) by simply 1, for any/ £ C 0 (G) 

/ A 1 € C 0 ((?) 

we are assured of the representation 

Uf) = [fdp 

J o 

where the integral is taken in the customary sense, f 
This concludes our discussion of the Daniell extension ap- 
proach and we will now turn toward a more measure oriented 
procedure. 


t Loomis £l], p. 35. 



160 


Element* of Abifraef Harmonic Analysis 


A Measure Theoretic Approach 

Before beginning please note that most of the pertinent 
terms used here are defined in the appendix First the genera! 
procedure will be outlined and then it mil be applied to the 
special case under consideration here, but before that wc 
will need the following 

Representation Theorem. f If J(J) is a positive linear func- 
tional on Ce(G), then there exists a unique Borel measure, 
n, such that 

JU) = [fd* 

J o 

Although this theorem will not be proven here, we will indi- 
cate how the (nght invariant) Borel measure is constructed 
from the / ( /) defined before W e first define for any E £ 6, 
the class of all compact seta in G, 

\(E) = inf /(/) 

/«a* 

Bs - \S 6 CJ(G),| j(g) > k M {g) for all g 6 (?) 

and ke is the characteristic function of E We state without 
proof that X is a regular content on C Any content however, 
induces an inner content as follows 
Denote the class of open sets of G by 0 and take, for 
06 0 

X.(0) * sup X(F) 

rco 

r«f 

Loosely, the "biggest*' compact set in 0 is desired here X« 
is now our induced inner content on 0 This done we can 
now define an outer measure a*, on G Suppose E C.G Take 

m *(E) ~ inf X.(0) 

XCOttf 


t For this result and for the rest of the chapter a good source is 
IfaJmoe C3J 



9. Right Invariant Haar Integral 


161 


Loosely, here, we wish to “approximate” E by an open set, 
Next we define the n*- measurable sets to be those sets, E, 
such that 

n*(A) = (A n E) + n*(A n CE) 

for any set A C G.f In other words the /immeasurable sets 
split all other sets additively. It can be shown that the n*~ 
measurable sets contain all the Borel sets, S, and it can 
further be demonstrated that when one restricts p* to S, 
calling the restriction fi, that a regular Borel measure is 
obtained. Diagrammatically the process is as follows: 

1(f) over C 0 (G) 

\ 

X over C 

\ 

X. over 0 

\ 

n * over all subsets 

\ 

n over S 

where the arrows are to be interpreted as “gives rise to.” 

Further the measure p so defined is right invariant by the 
following : 

Theorem. Let X be a locally compact Hausdorff space and 
let T be a homeomorphism where 

T: X->X. 

Then, if X is a content, 

X(F) = \(T(F)), F compact, 

is also a content and the corresponding measures have the 
property that 

fi(E) = n(T(E)) for all E £ S 


t Please note again that we are assuming the entire space, G, to be a 
Borel set. Otherwise. 1 , we would have to consider open Borel sets, and 
consider p* on the hereditary cr-ring of all <r-bounded sets. 



1 62 Element! of Abstract Harm ante Analysts 

where n is the measure determined X and p is the measure 
determined by X 

In our case we note that since I(/») = /(/), where 
M§) *= Sigh) then \(Eh) = \{E) for all E 6 C by the 
definition of A Consider now the homeomorphism 

T G—*G 
Q-*gh 

and consider X(fT) =■ \(r(£)) = A {Eh) = A (E) By the 
above theorem we can now say that 

ME) •= u{E) ~ n(Eh) 

where A —* n and X — * p which establishes the right invariance 
of n 

It is to be noted that the fact that A is a regular content 
guarantees for any compact set F that 

ME) = A (F) 

so that v is indeed really an extension of the A we started n Uh 
In summary \\c have obtained the following 

1 A Borel measure that is right invariant or equivalently 

2 a measure that is right invariant on all Borel sets 

3 Hence the corresponding integral is right mvnriant on 
all Borel functions Even though the following two facts 
have been implicitly established by the preceding discussion 
we shall prox e them here to be able to insure that our exten 
sion is a right invariant Haar measure 

(A) If 0 £ 6 then MO) >0 provided 0 ff 

(B) If E is compact then ME) < « 

Proof of A Let 0 € 6 0 0 

Since 0 9*0 there must be some element g € O Since G 
is regular (see exercise 4 of Chapter 6 and p 152) there is 
some neighborhood U of g such that 

gtVCO 

where V is compact Then, as we know, there must be a 



9. Right Invariant Hoar Integral 


163 


tinuous function/ such that 

0 </< 1 


1 where 

/ = 1 on U and / = 0 on CO. 
jarly we have 


ko(g)>f(g), all g. 
nv since / ^ 0 we have 


0 < 1(f) = f f dfi < (k 0 dti = „(0) 

J G J G 

hich proves A. 

Proof of B. Let E £ C. Recalling the proof of Theorem 4 
f the last chapter, there is some continuous function h € 
't(G) such that 

h = 1 on E. 

't is clear that h > k E , the characteristic function of E, But 
f hdn = 1(h) < <» . 

•'o 

Therefore 

»(E) = f k E dn < (hdti = 1(h) < «. QED 

~ G ^ G 


Appendix to Chapter 9 

Definition 1. A real-valued mapping, X, is said to be a 
regular content on C, the class of all compact sets of any 
topological spacef if 

1 . E e <?=> 0 < \(E ) < 

2. E h E 2 e C and E x C E 2 =>\(Ei) < h(E 2 ). 

t Although we can define a regular content on any topological Space, 
the most interesting applications arise when the space is a locally com- 
pact Hausdorff space. 



144 Clemen ri of Abstract Harmon c Analysis 

3 E x E t € d and E x ft F* = p=> X(£, U E t ) « X(£’ 1 ) + 
X(F,) 

4 FT, F, € £=> X,(F, U £7,) < \(F,) -f X(£7,) 

5 (Regulant)) Denote the set of interior points of a 
set F by F° and let £ € C Then 

X(F) = mf{X(F) \ E C F° C F € C| 

Definition 2 The induced tnner confenf induced by X X« is 
given by 

X*<0) « sup|X(F) | F C O, F e Cj 
where O^O 

Definition 3 let f} be a hereditary c ring The mapping 
n* fl — » 5 U | -f-°° } 
is •'aid to be an outer measure if 

(1) is* > 0 

(2) n, ■ H. i 

(3) (Monotonicity) If £i Ej € tl and E\ C Fj then 

h*(£.) < M*(&) 

(4) m*(0) - 0 

Definition 4 5a «ubsct of a topological space is said to be 
o-ixmnded if there exists a sequence of compact sets {F«|, 
such that 

BCU£, 


Exercises 

1 Pro\ e that e\ ery Borel measure is a finite i e if a is a 
Borel measure and if E is an) Borel set, then there exists 
s sequence {E n f o! Borel sots such that F d lf%}E» 
and n(E m ) < « 



9. Right Invariant Haar Integral 


165 


2. Let fi and v be two Borel measures such that v{B) = 0 
whenever y(E) =0. Prove that, if y is regular, v is also 
regular. (See footnote on p. 215 for the definition of a 
regular measure.) 

3. Let G be a locally compact topological abelian group. If 
n is a Haar measure on G, show that y{E~' 1 ) = n(E), for 
any Borel set E. 


References 

1. Loomis, An Introduction to Abstract Harmonic Analysis. 

2. Naimark, N armed Rings. 

3. Halmos, Measure Theory. 



cHAprfR ro 


The Daniell Extension from a 
Topological Point of View, 
Some General Results 
from Measure Theory, 
and Group Algebras 


In this chapter an alternative approach to extending the 
integral defined in Chapter 9 will be illustrated Essentially 
it is just the Darnell extension but from a topological point of 
view Next it will be shown that the integral defined over 
Co (G) m Chapter 9 is unique to within a constant multiplier 
and then some examples of right Haar measures will be given 
We will then sketch some general results from measure 
theory and state the Fubim theorems used in Chapters 1 
and 2 m the genera! situation Final!) we will include some 
discussion of group algebras 

Extending the Integral 

Suppose A is a locall) compact Hausdorff space and denote 
bj Co (A ) all those complex-valued continuous functions, /, 
over X with compact support £Wc note that Co(A') is a 
linear vector lattice such that, if / g 6 C, {X), then/ A 9 € 
C*(X) where as usual, / A g denotes the function defined 
as inf( f(x) g{x)) ] In addition, suppose L(f) is a positive 
linear functional defined on Co(A') fSmee L{ /) is a positive 
linear functional it immediately follows that 

(£(/)! < win 

Definition An extended real-valued function g is said to be 
lower semieonUnuous if for an) xa £ X and an> t > 0 there is 
ttfvgkbsrhaod of a ft {* 9 ) s nmJ? tint /©r^D -S' 6 

g{*) > oM - € 

J66 



10. The Daniell Extension 


167 


We will denote by M+ the set of all nonnegative lower semi- 
continuous functions. It can be shown that for any / £ M+ 
that 

/ = sup Y, = sup {<7 € Ct(X) \ g <f\. 

Putting this in a very loose way one might say that it is 
possible to approximate any lower semicontinuous from 
below by a continuous function. We now have our first ex- 
tension: for any / £ M+ we define 

L*(f) = sup L(g) 

and it can be shown that L* is almost a positive linear func- 
tional on M + ] namely, L* is additive, L*(cf ) = cL*(f) if 
c > 0 , and L* is positive. To extend this further let / be an 
arbitrary extended nonnegative function and consider 

Z f ={ge M+\g>f\. 

(Clearly Z t ^ 0 because g = «> is in Zj.) We can now 
define 

L,(/) = inf L*(g). 

BtZ/ 

Lt is unfortunately only subadditive, however (over the 
class of nonnegative functions) ; i.e., if /i, / 2 > 0, then 

L*(fi -h/2) < + L»(/ 2 ) 

in general. Actually L* is countably subadditive. Given L* 
though, we can define an outer measure, n*, as follows: 

Let A be a set and let k A be its characteristic function. 
Takef 

m*(A) = L.(/c a ). 

We will call A a zero set if ti*(A) = 0 . This done we can 
now introduce the following equivalence relation among the 
functions defined over X : 

/1 ~ /2 if and only if /i = / 2 everywhere except on a zero set. 


t It is left as an exercise to verify that A is actually an outer measure. 



168 


Element* of Abstract Harmonic Analysis 


Consider now the space of al! equn alence clashes of complex- 
valued functions,/, such that 

Mi/i) < * 

and call this class 1/ If w e take as a nonn on L 1 

ll/ll = MI/I). 

it can be shown that L l is complete with respect to this 
norm Also, by a previous observation, | L(/) j < (1/ 11 for 
/ C.(X) Consider now the closure of C t (X), C#(A),and 

call this class Li Li is a closed subspace of L l and, b> virtue 
of this observation, it follows that In is complete (A closed 
subspace of a complete space is complete ) Denotin g the 
continuous extension of Lover C#(X) to L.over C«(A) = Li 
w e w ill call a 6Ct A summable if and onlj if k A £ Li Using this, 
however we can actually define a measure, n, on the class 
of summable sets by taking 

/i(A) = L.(U 

We will call a 6et A, measurable if the intereclion of A 
with anj compact set is summable and it can be shown that 

/( L.U) > j (/* < » 

Conversely it can al«o be shown that 

j fd(t < «>=>/€ Li and L,( /) = j f J d m 

One can also show that all open and closed sets are meas- 
urable 

Diagrammatical!.} what ire have done can be summed up 
as follows 

(subadditiv e) L.(/) anj nonnegative/ 

I 

MHav* Jf + 

I 

L(/) over C.(X) 



10. The Daniell Extension 


169 


where the lines should be read upward as “has been extended 
to.” Using the L»(/) we then got 

L'(f) over C a (X) C L 1 
[via L.(/)] 

L{f ) over C„(X) 

where L r ( /) is the desired extension and completes our 
general outline of this extension procedure. 

In the special case where X — G is a locally compact 
topological group and L is right invariant on Ca(X), it can 
be seen readily that L e is right invariant on Co(X). 

We can now proceed to discussing the uniqueness of the 
/(/) [over Co(<?)D defined in the preceding chapter. 

Uniqueness of the Integral 

As usual, G is a locally compact topological group and 
whenever G is used in the remainder of the chapter it will 
be assumed to have those properties. In the preceding chapter 
the right-invariant Haar integral, 1(f), was defined over 
Co(G) and we now wish to suppose that another right- 
invariant Haar integral, J(f), has also been defined over 
Cii(G) . If we can show that ./( /) = al(f), where a is con- 
stant, on C b (G) then, .by any of the extension procedures, 
we can show that (denoting the extensions of I and / by I 
and J) J(f) = o.I ( /) on all Borel functions. To this end let 

/,HOT, v^O 

and choose d, and g< such that 

f(g) < for all g. (1) 

This implies that 

j(f) < 

' 1=1 

or that J(f) /J(f) is a lower bound for all 23>d>- Hence 

J(f)/JW < (}■■*) 



170 


Elements of Abstract Harmonic Analysis 


or 

J{f) < U *)J(+) (*) 

Consider some function, fi € Ct(G), such that ^ 0 Bj 
Theorem 5 of Chapter 8 there exists some neighborhood of 
the identity, T, such that for all ^ € fY 

t»(g) - l/i(ff) - TfsHsa.) I < 2»i (2) 

where <i is an arbitrary positive number, $ is assumed to bo 
not identically zero, and the c, and g, are as m that theorem 
Suppose now that »i < j and recall that there must exist a 
compact set, Ei, such that 

I g\ft(9) > 0 } c Bt 

Further (See proof of existence theorem in Chapter 0) all 
id's must vanish outside the compact set 
tM'r i £i u Ei 

where V\ corresponds to <i => J We can now assert the exist- 
ence of a function /* £ Cf(G) independent of <j such that 
(see proof of Theorem 4 Chapter 8 where such a function 
is mentioned) 

/, - l on iVf'EiUF, 

This and Eq (2) imply 

A(sr) + 2<./,(»> > O) 

and also 

Ma) S 2*1 ft(p) + £t*(w.) « 

I rom (3) it immediately follows that 

Jifi) + 2 .,/(/,) > < 3 «) 

while (4) implies that 

</»*)- &,(/, *) < 


(4a) 



10. The Danieil Extension 


171 


Multiplying through in (4a) by J(\f>) and using (4a) in 
(3a) we have 

J(h) + 2ftJ(A) > (l - (hU)JW- (5) 

Dividing (5) through by /(/) and using (*) we obtain 


J(h) 

JU) 


+ 2ej 


JU*) 

JU) 


> (1 - 2 £l (/ 2 :/,)) 


hUd 
hU) ' 


(6) 


Taking the generalized limit over t p first, and then letting 
ei — > 0, we have 


J(h) > iuo 
JU) ~ Kf) ' 

But, since / and fi are interchangeable we also have 


JU) > KR 
J(h) ~ Kh) 

which implies 


Kf) 



JU) 


for any / £ Cf(G), and, consequently, for all / 6 C 0 (G) . 
Calling the positive constant 1(h)/ J (h) — a we have 

/(/)=«/(/) for all feC»(G), 


or equivalently 


mi = a MJ 


where mi and mj are the right-invariant Haar measures deter- 
mined by I and J, respectively. This completes the proof. 



172 Semen tj of Abjtroct Harmonic Analysis 

Example* of Haar Measure* 

Example 1 ff, + («e the additiv e group of real numbers) , 
n Lcbcsgue measure 

right invariant integral j j{x) dx 

Example 2 R*, • (le the multiplies tn e group of non- 
zero real numbers) , n Take, for any Bore! set E C R*, 


J t x 


where the integral is taken in the Bebesquc sense To prove 
n is right-invariant consider 


X<«) - / 7 


dx 

d(xt) ( dx 


As nght invariant integral we can now take 

/ /(*) *(«> 

J R* 

Example 3 C* Take for any Bore! sct,£,in C*, 

/ dr 

where v represents two-dimensional Bebc«gue measure 

,,, . / '!• r ii.r<fr 

(Note that the Jacobian of the transformation is just | U ]’ 
here ) Hence m is nght in\ anant and n e have as a right- 
jmaxaat miegjsd 

f mdMd) 

J c • 



1 0. The Daniel! Extension 


173 


Remark. In each of the above examples the underlying 
groups have been abelian which implies that if p is right- 
invariant it will also be left-invariant. Further, in any 
abelian group we also have ti(E) = y.{E~') (by exercise 3, 
p. 165). In our next example a nonabelian group is con- 
sidered. 

Example 4. As the underlying multiplicative group G 
consider all matrices 

( x y\ 0 < x < oo 

0 1 / —00 <?/<«> 
and take as a topology the following: 

It is clear that a 1-1 correspondence exists between the 
points ( x , y ) of the open right half plane and all matrices of 
the form above or there exists a mapping 


x y\ i-i 


(x, y )■ 


As a topology on the space of matrices above we will take 
the associated topology from the plane. 

This done we will take, for every Borel set E in G, 


and v{E) = u 


dx dy 

E J X 


Now let 


£ = 


(a b\ 


VO 1 ) 


We note that this implies that the elements of %E are of 
the form 



where 


(x y\ 


VO 1 J 


6 E. 



174 ETemenfj of AbiJroef Hormomc Analyst 

In the plane this corresponds to 
x — » ai 
y -» ay + 6 
so 

-<® -/„/*? 

because the Jacobian of the transformation is just a 1 Hence 
(i is left invariant Consider £$ now The elements of /£ are 
of the form 



whieh corresponds in the plane to 
x -*ax 
y -* bx + y 

which has the Jacobian a Hence 

wU**- 

and w e ha\ e established the right im anance of r Consider 
now 

CD'-CT) 


and note that 



1 0, The Daniel! Extension 


1 75 


which corresponds to 


x — > 


1 

x 



Hence 

^ i} = fJl dxdy=i,(E) - 

It is clear now that there exists E such that 

n(E) < oo 
but 

v(J?) = m^' 1 ) = «. 

Our next theorem gives us an equivalent way of deter- 
mining whether a topological group is compact or not. 

Theorem. G is compact if and only if n(G) < a> for any 
right (or left) invariant Haar measure, p. 

Proof ( Necessity ). Suppose (7 is compact. This implies that 
the identity function, 1, has compact support or 1 6 Co(G). 
We now have 


m(G) « dp = 1(1) < =0. 

J a 

( Sufficiency ). Suppose (7 is not compact and let F be a 
compact neighborhood of the identity. Since the finite union 
of compact sets is itself compact then we must have 

G s* U Vg< 

i- 1 


for any positive integer p. 



J 76 Element of Abifroct Harmonic Anatyui 

Choose elements, gi, g%, now such that 

?.{ U Vg t 

i-i 

and let U be a symmetric neighborhood of identity such that 
IP C V We now claim that the sets {£/».} an; disjoint 
Suppose not, then (assuming that n > m) Ug, fi 
Ug m / 0 s=> there exist ui «, m U such that 

«i g, ~ «*?» =» g, = f IPg* 

But I Pg m C Vq<* Hence we have contradicted the very way 
in which the g K were chosen Therefore the sets \Ug.\ are 
disjoint Now wc have 

O O U Ug, 

which implies 

n(<7) S =* Iim » 00 

which completes the proof 

Remark If the above theorem applies then the Haar 
measure is usually normalized «o that n(G) *= 1 

Product Measure* 

Consider two sets X and Y and corresponding i rings 
of sets Sx and Sr Let ^ be a measure defined over Sx and 
v be a measure over Sr What we would like to do now w 
define a measure over some collection (<y-nng) of sets in 
A X r To this end consider 

Sj X Sr the a-nng generated by all A X B where 
/l € Sx and B £ Sr 

e would now like to state some results from measure 
theory but before doing so two new definitions are necessary 
Let f be a ret m the product space X X I , the set 

F, “ iy[{x,y) € Ft 



1 0. The Daniel! Extension 


177 


is called an x-sedion of E. Similarly the set 
E v = (*| (*, y) € E) 

is called a y-sedion of E. Consider some real-valued function, 
f(x, y), over X X Y now, and consider the function 

f x : E X —*R 

where f x (y) = f(x, y). The function f z is called an x-sedion 
of f. Similarly the function 

E v — » R 

where /„(x) = f(x, y) is called a y-sedion of f. We can now 
avail ourselves of the following results from measure theory. 
In the following statements X and Y will be sets, Sx and 
Sr will be <r-rings of sets in X and Y, respectively, and fi 
and v will be <r-finitef measures over Sx and Sy, respectively. 

1. If E € Sx X Sy, then every E x € Sy and every E v € 
Sx- We can reword this as follows: if E is measurable in 
X X y, then every section of E is measurable. 

2. If f(x, y) is a measurable function, then every section 
of / is measurable. 

3. If E is a measurable set in X X Y, then the functions 

p{x) = v{E x ) and q{y) = y(E v ) 

are measurable functions over X and F, respectively. 
Furthermore 

f v{E x ) = f p(E v ) dv(y). 

•' X Y 

4. Let E be a measurable set in X X Y. Then 

HE) = fv(E x ) dp(x) = f HE,) dv{y) 

J X J y 


t In this context a measure over a ring of sets, S, will be called tr-finite 
if for any set E £ S there exist E n € S, n = 1, 2, . - such that E <= 
Un-i E„ where (S„) < for all n. 



178 


Element* of Abitract Harmonic Anolyjlj 


constitutes a a finite measure on S. r X Sy and if A (■ Sx 
and B € Sr, then X(A XB) * n(A)v(B) 

The measure X is called the product mcajurc and is often 
denoted by X = n X * One also notes in passing that the 
propertj 

X(A X B) = *(A)»(0), X € Sx, B € S r 

determines X uniquely over Sx X Sr 
Having defined a product measure we can non talk about 
integrals over measurable subsets of A' X Y In the same 
notation as the preceding we shall call the integral 

f h(x y) dX « f hix, y ) d(p X >) 

J xxr J xxr 

a double integral Just as in the case of double or tnplc inte- 
grals in euclidean space we turn our attention next toward 
titrated integrals To this end consider 

f(x) *= f h(x y) <My) - / My) <M«/) 

J r J r 

assuming of course that the integral of My) exists If 
Jx f(x) dn also exists wc wntc 

f fix) dp = f f hix y) drdu = f duf h(x, y) dr, 

J x J x J r J x J r 

and call this an iterated integral We can, of course, inter- 
change the roles of X and F in the above discussion and 
consider 


( dr f hix, y ) d,i 
J r J x 

assuming it makes sense \\ ith these v cry natural generaliza- 
tions of concepts famifiar in euclidean space in mind, we 
can now state the generalized v eroons of the Fubim theorems 



TO, The Darnell Extension 


179 


we made such extensive use of in Chapters 1 and 2.f To 
simplify matters later on when we wish to refer to these 
theorems we shall call them simply the Fubini theorems. 
The theorems overlap in content, and we state them all 
simply for convenience. 


Theorem (FI). If h > 0, and measurable on X X Y, then 
I h dk = I I h dv dfi = I j h dy dv. 

•'t-W j y J v J v J y 


Theorem (F2). If h is integrable on X X Y, then almost 
every section of h is integrable which implies 

f(x) = [ h(x,y)dv(y ) = f h x (y) dv(y) 

J r J y 

exists a.e., and also that 

g(y)~ = / h(x, y) dy(x) = / hy{x) dy{x) 

J x J x 

exists a.e. Also /(a;) and g(y) are integrable, and 

f fhd\ = f j h dv dy = j j h dy dv 
•'wv * J x ' Y •'* v •'Y 


'XXY 1 


Theorem (F3). If h is measurable on X X Y, and one of 

f | h | dX, f f | h j dv dy, f f \ h\dydv 
J XXY " v “ v ^ v ^ Y 


’ Y ’'X 


exists and is finite. Then all three exist, are finite, and are 
equal, and 

f h d\ — f f h dv dy = j J h dy. dv. 

■'xxy •'x-'r J Y J X 


t See Theorems 3 and 4 (Fubini’s theorem and the Tonelli-Hobson 
theorem) of the appendix to Chapter 1 on pp. 16-17. 



1 80 Element* of AbUract Harmonic Analysis 

Remark It can be shown that any Borel measure is <r- 
fimtc (see exercise 1, Chapter 9) and, further, that if n and » 
are two regular (see footnote, p 215) Borel measures over 
the Borel seta Sx and Sr, respectively, then u X r is also a 
regular Borel measure 

Suppose now that X and Y are locally compact Kausdorff 
spaces and that ft and r arc regular Borel measures defined 
over the classes of Borel sets S x and Sr, respectively, where, 
of course, Sx is a class of subsets of X and similarly for Sy 
By virtue of the remark, tn any framework of this type, 
fi X v must be a regular Borel measure With this in mind 
consider a locally compact topological group, G, and a right 
Haar measure, n Denote by 

Li(Q) all equivalence classes of complex-valued Borel 
measurable functions, /, such that f 0 \f\d>i < « 

As a norm on U(G) we can take 

ii/ii, - / m* 

We can show that L\(fi) is actually a Banach algebra with 
respect to this norm if we take as multiplication / * g, where 

(/•?)<*) *= / /(* y~ l )g(y) My), f,g £ h(G) 

J a 

In order for the above definition of multiplication to make 
sense, however, we must show that the integrand is measur- 
able, and that the integral exists for almost every z In order 
that Lj{G) be a Banach algebra we must then show that 
/•g € Bi and l[i < ||/jh jiff lit* and we shall proceed 
to the demonstration of all these facts now 

Let 0 be an open set in the complex plane and let E ** 
/“•(O) Denote f(zy~ l ) by h(x, y) If we can show that 
k-'fQ) - |<ar, y ) \Jfzy~') Z 0 or zjr' Z f'HO) - E) 
is a Borel measurable set, then we will have established that 
h is a measurable function and, since t c(z, y) ** q(v) w 



10. The Daniell Extension 


181 


clearly Borel measurable, we will have that the integrand, 
is a Borel function on G X G, and thus every section is 
measurable. 

It is clear that the set Ei = E X G is a measurable set in 
G X G. We would like to show that 

h~ l (0) = E 2 = [(x,y) 6 (? X G | xy r 1 6 E j 

is a measurable set in G X G. 

To this end we note that since homeomorphisms preserve 
compact sets and, hence, Borel sets, that if we can show 
Ei to be a homeomorphic image of E x we will have estab- 
lished the measurability of E 2 , because we have already 
noted that E\ is measurable. Proceeding according to this 
plan of attack consider the following homeomorphismt: 

G X G G X G 
(x, y) ->• (xy, y) 

Let (x, y ) 6 Ei. By the definition of E%, x 6 E and y 6 G. 
Now since xyy~ l ~ x € E it follows that (x, y) has its image 
in E 2 . Conversely suppose 

(x, y) € E 2 => xy~' € E=> (xy- 1 , y) 6 Ei. 

But 

(xy- 1 , y ) -> (xy- x y, y) = (x, y) . 

Hence for any point, (x, y), in E 2 we found a point in E t 
that maps into it. Thus we have established the fact that h 
is a measurable function, and also the measurability of the 
integrand in question. To show that the integral exists for 
almost every x, consider 

f dy(y) f \Kxy- 1 ) 1 1 g(y) | dy(x) 

J G J G 

= f \g(y)\dn(y)f {/(xy- 1 ) \dy(x). 
J a j g 


t It is easily verified that this is a homeomorphism by using Axioms 
G3 and G4 for topological groups listed in Chapter 6. 



1 82 Elements of Abstract Harmonic Analysis 

Using the right invariance of » m the last member wc ha\e 
that the above product is equal to 

ll/HilUII. 

By Theorem F3 of this chapter, 

[ /tar‘)ff(y) d(n X n) 

j gxg 

exists and is finite Also by Theorem F2, 

[f(xy-')g( v) My) 

J o 

exists for a c x and is integrabic Therefore / • g 6 1* 

Now we must show that 

11/ *Pl!i 5 11/ Hi 11 9 jit 

The existence of 

II/* fill - / I (/•*>(*) I <W x) 

J a 

has been established Expanding this wc have 

11/ • g Hi = f <M*) I f f(xy~')g(y) My) I 

J o J a 

< I dp(z){ |/(iy-')p(y) l<fr(tf) 

J a J a 

Interchanging the order of integration in the last term we 
have that it is equal to 

I 1 ff(y) I ifci(y) j i/(r tr l ) I <&»(*) 

J o J tj 

which, bj the right invariance of u equals 

ii/ ii> iip ii. < - 



1 0. The Darnell Extension 


183 


which justifies the above interchange in the order of inte- 
gration and establishes the desired result. Thus L X {G ) is a 
Banach algebra. 

Theorem. Let G be commutative then f *g — g* f. 

Proof. By definition 

(f*g)(x) = J f(xy~ l )g(y) dy(y). 

Replacing y by yx in the integral and using the right invari- 
ance of u we have the above integral equal to 


ff(y~ l )g(yx) dy(y). 


Replacing y by y~ l and noting that G being abelian implies 
that n(E) — n(E~ l ) we have 


U*g)(x) = J f(y)g{y~ l x) dn(y) 

= Jg(xir l )f(y) dy(y) 


“(»*/)(*). QED 

It can also be shown that the converse of this statement' is 
true (see exercise 1). 

An interesting reflection of whether an identity is present 
in L X {G), in the topological character of the underlying 
group follows. 

Theorem. Li(G) has an identity, if and only if G has the 
discrete topology. 

Proof. Suppose first that G is discrete. In this case the 
one-point sets of G are open sets, each of which have the 
same positive measure, which we may assume to equal 1. 



1 84 Element! of Abstract Harmonic Analysts 

Symbolically, if ne denote the nght Haar measure by n 
and let g be an element of G we have 

v{\o)) - 1 

This convention then implies 

//<*) *(*) - Em 

■'o icO 

which means that /is integrable if and only if / *= 0 cieiy- 
where except on some countable set Xi, x t , • • • and 

52 l/(*.) I < « 

Now consider 

(/•ff)(*) - fj(xy~ l )g(v) dn(y) 

- 52/(*jr‘)y(y) 

ytO 

and take 

1, x ■= e 

e(x) >= 

0, x He 

where e is, as usual, the identity clement of G Wc now ha\ e 
(e •g)(x) = 52*fo r‘)p(y) - p(*) 

and can, in a similar fashion show that ( g • t) (x) — g(x) to 
establish the sufficiency of the condition 

Convene Suppose there exists an identity, m, for L%{G) 
We now contend that this supposition implies that the set 
of all measures of nonempty open sets in G has a positive 
greatest lower bound Suppose not, this means that guen a 
8 > 0 there exists some nonempty open set, Ui, such that 



10. The Daniel! Extension 


185 


n(U s ) < S. Since U s is nonempty there must exist some 
element g € Us- By the right invariance of n then we have 
that n(Us) = But, since e 6 Uig~ l , we have 

established the existence of an open set, 0 = C/jjr 1 , contain- 
ing e such that n(O) < <5. Now since one can show that 
C 0 (<7) is dense in Li(G ) , given any £ > 0 there must be a 
function/ 6 C 0 (G) such that 

f | m - f | d,i < .f. 

J a 

But 

f \m\dn < f \m - f\d/i + f \f\dfi 
J a J a J a 

and in general, for an arbitrary measurable set E, 

[ |/| dn < max/ • n(E). 

J E 

Hence, since we are unrestricted in the choice of S, it is 
clear that we can choose an open set 0, containing e, such 
that 


/ | m | d/i < £ 

■’o 

for any £ > 0. With t given and an appropriate 0 chosen 
we now select a symmetric neighborhood of e, V, such that 
V 1 C O.f Denoting the characteristic function of V by kv 
we have 

k v (x) = (kv * m) (x) = / kv(xy- l )m(y) dv(y) 

J a 

for a.e. x € V. 


t The existence of such a V is guaranteed by observations 10 and 11 in 
Chapter 6. 



10. The Daniel) Extension 


187 


3. Let 1(f) = fa f(x) dp be a right invariant integral on 
G. Let J(f) = faf{yx) dp. Show that J is also a right 
invariant integral, and, hence, J(f) = A(y)I(f). A (y) 
is called the modular function of G. Prove that A (y) is a 
continuous homomorphism of G into R. 

4. Using the notation of exercise 3, show that if G is compact 
or commutative, then A (y) = 1 for all y 6 G. 

5. Let I be an integral on Co(X) , where X is a locally com- 
pact Hausdorff space. If / € C<>(X ) , define ||/|| = 
supxex |/(x) |. If E is any compact subset of X prove 
that there exists a constant k = k( E) such that 1(f) < 
fc |[/(| for all / € Co(X ) provided [x |/(x) ^ OJ C E. 

References 


J. Naimark, Normed Rings. 

2. Loomis, An Introduction to Abstract Harmonic Analysis. 

3. Rudin, Fourier Analysis on Groups. 



CHAPTER 11 


Characters and the Dual Group 
of a Locally Compact, Abelian, 
Topological Group 


In this chapter we shall continue the discussion of the 
Banach algebra L}{G) that we introduced in the last chapter 
Next we shall turn our attention toward the concept of a 
character and of the dual group of a locally compact, 
abelian topological group Some examples of characters in 
familiar frameworks and certain theorems pertaining to 
characters and the dual group are then discussed 

Remark It is easily verified that Li(0) is an algebra with 
involution if we take for/ 6 Li((?),/*(x) = f(xr l ) 
Throughout this chapter we shall assume that G, in addi- 
tion to being a locally compact topological group is also 
abelian As noted in the last chapter L } (G ) is an algebra 
with identity if and only if G is discrete In any case by the 
results of Chapter G the Banach algebra Li(G) can be ex- 
tended to an algebra with identity which we shall denote by 
R(G) hence /?((?) is a commutative Banach algebra with 
identity which we shall denote by e f To facilitate matters 
in the discussion that follows we introduce the following 
notation Let / € Li(G) \\ e now define/* y) to be 

My) = /(v*- 1 ) 

We now contend that the mapping 
X -»/, 

is uniformly continuous 


1 HcVe toaX we to toe xised Wi Ctiaptasa S *®d. 

4 insofar as the multiplicative identity of the Banach algebra I* being 
denoted by e 


188 



1 1 . Characters and the Dual Group 


189 


Proof. Let / € Li(G) and let e > 0 be given. Since C 0 (<?) 
is dense in Li(G) there exists a function g £ C 0 (G) such that 
11/ — g l|i < e/4. Since g € C 0 (G), however, this implies 
that there exists a compact set, E, such that 

g(CE) = 0 

and that g is uniformly continuous by the first theorem in 
Chapter 7. Hence there must exist some neighborhood of the 
identity of G, V, such that 


\g(y ) - 

aiyx- 1 ) | 

6 

< 4p(£) 

for 

yiyx- 1 )- 1 

or 





1 ff(v) • 

- g*(y) 1 

€ 

< 4 y(E) 

for 

x 6 V. 

Consider 

now 

• 





II g - g* 

Hi = / 1 

J G 

g(y) - 

g*(y) 1 dn(y) 

Since g x (y) vanishes if yx~ l 

$ E we 

can write 


II g - g x !li = [ I g(y ) - 9*{y) I 

J E\jEx 

If x € V, we have 


which implies 

11/ - /- Hi < 11/ - g lit + il g - g* Hi + II fc - U Hi 

< 7 + T + 7 = e for x e V '^ 

4 4 4 


( 1 ) 


t Note here that the right invariance of the integral implies that 
11/ -Plli - |l/. -fcllt. 



190 


Eementi of AbitrotS Homcnc Anctyuj 


We now claim that 

/* -/» • /-Ar. (2) 

because 

(ft - /•) (z) =/»(*) -f,(z) - /(z^ 1 ) - /(fir 1 ) 

while 

(/-UW = (/-A-K*r*) 

»/(£T J ) -fizz-hy-*) 

« /(er~«) - 

which demonstrates that their effect on any group element, 
z, 13 the same and proves their equahtj In light of this 

ii/.-/, ii. = ii(/-/^)«n. 

which, by the nght invariance of the integral, is equal to 
'I (f - fm) Hi 

But, as eh<3wn by (!), 

11/ - /*r* Hi < * t or ipr x € 1 , 

or, using (2), 

HA - /. Hi < < for yr->0 (3) 

which is the *tatement of uniform continuity and completes 
the proof 

Our next theorem is concerned with an approximate 
identity for Lj{G) (See Chapter 1, p 9, where the notion 
of approximate identity is fir^t mentioned ) 

Theorem 1. Let / £ Lt{G) and let «> 0 be gnen Then 
there exists a neighborhood of the identity of G, I , *uch 
that if ii is a nonnegatne Borel function which vanishes 
outs’de V, with the added propertj that 

f u(z) dn(z) = 1, 

J G 

then |j/ — / »tt |ji < < (u is an approximate identity) 



1 1. Characters and the Dual Group 


191 


Proof. By the preceding result, with x taken to be the 
identity element of G in (3) , we can assert the existence of a 
neighborhood, V, of the identity of G such that 

11/ — /v (fi <« for y e V. (4) 

Let u satisfy the hypothesis and consider 

( f*u)(x ) -J{x) = f ifixy- 1 ) - f(x))u(y) dyiy). 

J G 

We now have 

||/*M-/|li= [ dy(x) \ [ if(xy~ l ) - f(x))u(y) dy(y) 

J G I J G 


< [ dyix) [ | fixrr 1 ) - Six ) | • I uiy) \ dy{y) 

while interchanging the order of integration (which is 
justified by the final answer) yields 

I! / *u ~ / ||i < J I uiy) | dyiy ) J ! /„(*) - fix) | dyix) 


or 


ll/*«-/||i< f uiy) || / v - fWidviy). 

J G 

But, since u vanishes outside V , the last term is equal to 

f uiy) || / v - fWidyiy). 

J v 

Using (4) this must be less than 

« / uiy) dyiy) 

J v 

which, using the hypothesis about u, is equal to just e or 
||/ *u -/||i < « 
which completes the proof. 



1 92 Elements of Abstract Harmonic Analysis 

This next theorem has a certain esthetic appeal as well 
as being a rather useful result about any homomorphism 
mapping a commutative Banach algebra with identity into 
the complex numbers 

Theorem 2 Let X be a commutative Banach algebra with 
identity, e, where it is assumed that the norm of e is unity, 
and let / X — ► C be a nontrivial (therefore onto) homo- 
morphism Then JJ/|{ = 1 

Proof Let x € X and let f(x) = X Since / is nontrivial 
we immediately have /(e) = 1 , hence 

/(* - Xe) = /(*) - X/(e) «= X - X = 0 

which implies (x — Xe) 6 M, where M is the kernel of / 
Since / is a homomorphism onto C, AT must be a maximal 
ideal Therefore, by the first theorem in Chapter 4, x — Xe 
must be singular which is equivalent to saving that X € <r(x) 
By part (7) of the last theorem in Chapter 3 we have 

IM < II *11 <=*!/(*) I < 11*11 for any x € A 
which implies 

ii/ii<i m 

By the definition of [(/ j| we have 

!/(*) I <11/1111*11 for all zZX 
so, letting x = e, 

l - !/(*)! <11/11 II e II *■ 11/1) 

Combining this with (G) completes the proof 

Corollary Under the same hypothesis as the theorem, / 
must be continuous 

Characters and fhe Dual Group 

11 e now proceed to the notion of a character of a group 
a notion that will play an important role throughout the 
rest of this chapter and the next chapter 



1 1. Characters and the Dual Group 


193 


Definition. A mapping 

x: G->C 

with the properties that 

1. | x(s) | = 1 for all x £ G, and 

2. x(zy) = x(x)x(y) is called a character of G. 

We will denote by r the class of all continuous characters 
on G. If we take 

(xixi)(ff) = xi(ff)xs(ff) 


as the law of composition, it is easily verified that r is a 
group and is called the dual group of G. Denoting the exten- 
sion of L\{G) to a commutative Banach algebra with identity, 
e, by R(G) we wish now to determine the set of maximal 
ideals in R(G). First we shall demonstrate that Li(G) is a 
maximal ideal in R(G), but before proceeding we ask the 
reader to recall that a typical element of R(G) is ae + f 
where a is a complex number and / £ Li(G). With this in 
mind if we suppose that I is an ideal that Contains Li((?) 
we note that this would imply the existence of an element 
ae + /, a ^ 0, which is in I but not in Li{G) . Since / is an 
ideal, 


which implies 


{ae + /) — / = ae £ 1 
e £ I 


and, hence, I = R(G). Therefore Li(G), which is clearly 
an ideal, is a maximal ideal in R{G). Having gotten this 
one maximal ideal we will obtain the rest through the follow- 
ing theorem. 

Theorem 3. Let x be any continuous character of G, take 
/ £ Li(G), and consider 


fix) = Jf(x)x(x) dp(x). 



194 


Element* of Abslroct Hormonic Anolyju 


Then the mapping 

R(G)~>C x fixed, 

Xe + / — ► X + J(x) !> + / varying 

is a homomorphism of R(G) onto C that does not vanish on 
all of Li(G) Conversely, every homomorphism that docs 
not vanish on all of Li(G) is obtained in this fashion Further- 
more distinct characters induce distinct homomorphisms 

Remark Theorem 3 establishes a one-to-one correspondence 
between the dual group, r, and the class of all homomor- 
phisms that do not vanish identically on Li{G) Since there 
exists a one-to-one correspondence between the class of 
maximal ideals of an algebra and the set of all homo- 
morphisms t the theorem also establishes a one-to-one corre- 
spondence between the maximals ideals excluding Lt(Q) 
and the dual group 

Proof J et x 6 r and consider the mapping of 11(0) onto 
C mentioned in the statement of the theorem It is immedi- 
ately clear that the mapping is a linear functional In light 
of this we shall now demonstrate that the mapping preserves 
products to establish the fact that it is a homomorphism 
To this end consider the product! 

(Xe + /) (^e + ff) ~ Xpe + Xy + ft/ +f*3 

which, under the above mentioned mapping, maps into 

Xu + Xp(x) + m?(x) + /(/ •g)(y)x(y) d**(y) 

•’o 

t See m this connection the mapping mentioned in Chapter 3 1 » the 
theorem following the introduction of the quotient algebra 

t The reader should clarify for himself exactly what is meant by tins 
product of elements in R[G) 



1 1 . Characters and the Dual Group 


195 


which is the same as 

V + \g(x) + /i. f(x) + [ x(y) dfi(y) f f(yzr 1 )g(z) d/x(z). 

J Q J G 

(•) 

Interchanging the order of integration in the last term and 
noting that, since x is a character, x(y) = xfzjxO/z"" 1 ) , the 
above integral becomes 

f x(s)g(z) dy(z) [ }{.yz- l )x{yz-') dy(y). 

J G J G 

But, since the integral is right-invariant, ( *) becomes 

Xp + \g(x) + yf(x) + §(x)f(x) - (X + f(x))(y + g(x)) 

which, since we have already noted that the mapping is 
linear, establishes the homomorphism. We must now show 
that there is at least one / € Li(G) that does not map into 
zero under the above mapping. To this end take k £ Li(G) 
such that 


[ h dn 0 . 

•’a 

As the required / we can now take 

/ = h/ x 

it being clear that we are never dividing by zero because x 
is a character. Hence we have now shown that the homo- 
morphism does not vanish identically on L\{G) and, also, 
by the same argument, shown that the homomorphism is 
nontrivial and, therefore, onto. We can now proceed to 
showing that, given any homomorphism, there exists some 
character that “generates” it in the fashion illustrated above. 

Suppose now that h is a homomorphism mapping R(G) 
onto C that does not vanish identically on Li(G), and con- 
sider the restriction of h to just L\{G) ; i.e., take 

hi = h It, ((;>• 



196 


Elements of Abstract Harmonic Analysis 


By the preceding theorem we have that |j A || *= I which 
implies 0 < || Ai |{ < 1 Now, since G is a locally compact, 
abelian, topological group, the dual space of £i(<7) is iso- 
morphic to t and, since || Ai)| < I, one can, by using 
a well-known representation theorem, write 

hi (/) - M*) 

for some $ 6 L„{G) Furtherznorcuehave || |L = |1 h j) < 
1 Bearing these results in mind Jet/, g E Li(G) and consider 

/ m /)*(*)♦(») My) - Mf) f g(y)My) Mv) 

- hi(/)Ai(ff) (7) 

which, since /, g € Liifi) and h x is a homomorphism on 
L\(G) t must equal 

M/'ff) = /(/•(>)(*)*(*) Mx) 

- / ♦(*) Mx) { f(xy~')9(y) My) 

t Suppose X is a set S is a nng of Bets in A' and measure on 
S Now let £ £ S In this context L m (E, u) *» L m (E) consists of those 
functions / such that |/(z) | < M ae on E Denoting the class of all 
M that bound |/(r) | a e by M * it can be shown that the esimtud tup 

off, 

inf M — css eup/ ~ |1/||« 

i» actually a norm on the space L m {E) Furthermore it can also be 
shown that for/ € L m {B) 

toll/ll, - 1«4 / ~ *UP/ 

r-® / 

It ia assumed of course, that p(E) < « in the above discussion 



1 1. Characters and the Dual Group 


197 


Interchanging the order of integration this becomes 


Jg(y) dfi(y) j 'P{x)f{xy~ l ) dfi(x) 


= J g(y) dfi(y) J 4'(x)f v (x) dfi(x) 


= f 9(y)hi(f v ) dp(y) 

or, combining the first and last terms of this equality, 

/ h U)g(y)'Ky) dy(y) = J hi f v )g(y) dn(y) 

G J G 

for any g 6 Li(G), which implies 

hi(f)Hy) = hi{f v ) a.e. 

Now, by the first result in this chapter, / is a continuous 
function of y. By the corollary to the second theorem to 
this chapter, hi is a continuous function and we can now say 
that the composite function hi(f v ) is continuous. Since h, 
by hypothesis, does not vanish on all Li(G'), we can choose 
an / £ Li(G) such that fh( f) ^ 0 and can assert that the 
function 


'f'iy) = 


hUA 

hi(f) 


is continuous almost everywhere. We can now define /(y) 
to be continuous everywhere because this will not disturb 
our representation theorem [namely, that 


h(f) = jf{x)4>{x) dn Or)]; 

i-e., we will take 


tty) = 


hi(fv) 

hi(f) 


(8) 



1 98 Elements of Abstract Harmonic Analysis 

everywhere Using (8) and noting that = (/,)» we have 

- m/w> = «(/,).) 

- h(f.My) - hUlHxmv) 

and can conclude that 

*(*y) - tK*Wy) (9) 

As noted previously || p [[„ < I which implies 

!*(*)!< 1 ae (10) 

Since p is continuous however if | p I > 1 at x a then there 
must be a w hole neighborhood b of r» such that \p \ > l for 
all * € U Since n(U) must be positiv e this contradicts (10) 
Hence 

!*(*) I < 1 

everywhere From (8) we have the fact that tK*i) « I 
where ei is the identity of G Using this and (9) we have 
for any x 


Combining this result with (10) gives 

| p(x) | = I for all x 

Thus p is a character on G and we have expressed h t (/) in 
the required way It now remains to show that h can be 
expressed in this fashion Jvote that in the previous notation 
we now have 

MS) -SiP) 

I^et Xe + / € f?(G) and consider 

*(**+/> ->*<«> + MS) 

Since h is a homomorphism fc(e) *= 1 and we have 
+/) - * + /«■) 
which is the required representation for h 



1 1. Characters and the Dual Group 


199 


To show that distinct characters induce indistinct homo- 
morphisms let xi and xi be distinct characters from r, and 
suppose that, for any / £ Li(G), 

X + /(xi) = X +/(»). 

This implies 

f f(x)xi(x) dfi(x) = ff(x)xi(x) dn(x) 

J G J G 

for any / £ Li((7). Hence 

Xi = X 2 a.e. 

If two continuous functions are equal almost everywhere, 
however, they must be equal everywhere, which implies that 

Xi = X2 

and contradicts the assumption that xi and * 2 were distinct 
characters, and proves that distinct characters induce 
distinct homomorphisms. QED 
Having established the 1-1 correspondence cited in the 
remark and noting the 1-1 correspondence between the class 
of all homomorphisms and the maximal ideals, we wish to 
rewrite some of the results obtained here in terms of the 
notation used in Chapter 4. 

Let h be a complex homomorphism on R(G) and let M 
be the corresponding maximal ideal. Then, in the notation 
of Chapter 4 we have for (Xe +/) £ R{G) 

H\e + f) = (Xe + /) (M) 

= X +f(M) 

— X + hi( f) 

= x + fix) 

= X + f f(x)x(x) d»{x). 


(ID 



200 


Element* of Abstract Harmon e Analysis 


Similarly we can rewrite (8) as 


x(*) 


MM 

W) ’ 

f.W) 

S(M) 


for / such that h t {f) 7 * 0 


(/(M) * 0) (12) 


We would now like to show that any x 6 T in addition to 
being continuous must also be uniformly continuous and the 
notation used above will be helpful in proving this 


Theorem 4 Let x be a continuous character of G Then x 1 3 
uniformly continuous 


Proof Consider 


I x(i) - x{«) I 


i/.w - um 1 

l/<M) 1 


B> the last theorem in Chapter 3 we have 

~/,WI < II/, -A Hi 

iron 1 “ i/wi 

It now remains on)> to apply the first result obtained in 
this chapter to complete the proof 


Examples of Characters 

Example 1 I-et G be the additive topological group of 
real numbers It 4- and consider the conditions imparl on 
any character x namely that for ti tt (: R 

x(h + k) = X(ti) + XM 

and 

I x(0 1 = 1 for t e R 
A solution is offered by taking 

x(f) ** t i,r where x 6 R 

and thus a character is obtained for each x € R Actually, 
these exhaust all continuous characters of R Now consider 



1 1. Characters and the Dual Group 


201 


Li(G) and its extension, R(G). Using the notation of Chapter 
4 we obtained Eq. (11) which, in this case, becomes 

\+f(M) = X + nf(t)e«*dl. 

— co 

Since x is characterized by x we can write 

KM) = /(x) =f(x) = fme^di 

* —CD 

where x is the x corresponding to x- 

Example 2. Let G be the additive group of integers (with 
the discrete topology) , Z, +. Since this is a discrete structure 
Li(G) has an identity present and we need not bother with 
extending it. We wish now to obtain solutions to 

x(«i + rw) = x(«i) + x(« 2 ) («i, E 2) 

and 

! x( n ) | = 1 for all n 6 Z. 

Analogous to the preceding result we note that 
x(n) = e' na (a € R ) 

does the trick. 

We now let/ € Li(G) and proceed in the exact same fashion 
as the preceeding examples to get [using Eq. (11) againj 

KM) — fix) = f,Kn)e *- 

n~~ca 

and note the 1-1 correspondence 

{ X } e ia ( a mod 2 tt) 

where (x) is the class of all characters. 

We wish now to focus our attention on the topological 
aspects of r. 

Suppose M represents the set of all maximal ideals in 
R{G) and let Mo denote the maximal ideal Li(G). We have 
then the 1-1 correspondence 

M - {Mo} «-*■ r. 



202 


Element* of Abstract Harmonic Analysis 


With respect to the topology defined in Chapter 5 for Af 
it was shown that Af was Hausdorff and also compact (see 
first theorem in Chapter 6) Since, in general, removing one 
point from a compact Hausdorff space j lelds a locallj com- 
pact space (considers g , removing one point from a closed in- 
terval) we have that Af — JAfo} is a locally compact, Haus- 
dorff space We will topologize r now by taking the open 
sets to be those which correspond to open sets in A/ — |A/«) 
under the 1-1 correspondence and, with respect to this 
topology, can immediately assert that r is a locally compact, 
Hausdorff space To indicate more exactly what the corre- 
sponding topology in T is, we will exhibit a particular member 
of the fundamental system of neighborhoods in r To this 
end let xi be some particular element of T, let Afi he the 
corresponding element in A? — {Af#}, and consider the 
following n elements of R(G) 

gt = 4- /*, A, » 1, 2, •••, n 

Then the neighborhood about xi, denoted by W(xufi ■**» 
t) is given by all those x such that 

|<fc(A/) - | - \MAI) -M*h)\ 

= | Jfk(x){ x (x) - xi(i))*(x)| <« 

for k = 1,2, •••,» 


(Note that in the above expression the Af that appears in 
the inequality is the Af that corresponds to the particular x ) 
We shall presently prove that T is actually a topological 
group Since it has already been demonstrated that r is a 
group and that r is a Hausdorff space, it remains only to 
verify that the mappings 


and 


X -* X“‘ (x € r) 

(xi» xj) — * Xtx* (xi. xi € r) 


are continuous 



203 


11. Characters and the Dual Group 

„ .i i, we have obtained a topology for r, as it 
happens there is a slightly nicer characterization and this next 
theorem is needed in obtaining this topology. 

Theorem 5. The mapping 

G X r — > C 
(x, x) x(z) 

is continuous. 

Proof. Consider Mi ^ Mo- Now take (for / 6 U (G)) 

| / xl (Mi) - /*,(M) 1 < ~ 1 

+ \f xi (Mi) - fxi(M) |. 

By the last theorem in Chapter 3 [part (7)] we have 

\f xl (Mi) - fx t {Mi) 1 < \l/-i " 

which, by the first result in this chapter, can be made a rbi- 
trarily small for 6 W (for some neighborhood of the 

identity IF), whereas 

| /..(MO - /»W I 

. can be made arbitrarily small, less than . say, ^ “"“j 
M to 7 (Mi /„, «). [See Chapter 5 for the definition ox 
V{M h f xv e).] We have now established the continui y 

the mapping „ 

G X M - {Mo} -> C 

(x, M) —*fx(M) 

where /is a fixed element of Li(G) '• v that 

It is easily seen that there is some / € i ( 

/(Mr) X 0. 

At any point (xi, Mi), then, /(M) ^ °/r ,^//(M) is a con " 
neighborhood of Mi-t Consequently M argument, 

tinuous function of x and M by the preceding argu 

t The reader is asked to recall now that the top <=> ^ functions 

in Chapter 5 was the weakest topology that would make 
x(M), x fixed, M varying, continuous. 



204 


Elements of Abstract Harmon e Analys 


This m turn implies bj Eq (12) that the function of x 
and a: x(*) 


X(*) 


MV) 

fon 


is continuous It is to be noted that given any M there is a 
corresponding x and this is the x that is meant above This 
completes the proof 

W e now w i«»h to determine another basis for r s topologj 
It is first noted that denoting the fundamental stem of 
neighborhoods on r by | II ) since the 11 form a fundamental 
system of neighborhoods for r that if we can construct 
another sj stem of sets JA ) such that any 11 contains some 
% then clearly the JAT) form a fundamental system of 
neighborhoods for T also 

Let E be a compact set in G xo € I* t > 0 be given and 
take 

A'Xxo B t) - (xllx(x) - xi(z) I < « for all* € E\ 

It is first observed that this set is open by a theorem in 
Chapter 5 t where for the sake of comparison with that 
theorem we have 

GXT->R 


(x x) -* i x(x) - xo(x) I 

We will now show that the sets A (xo E «) form a funtla 
mental system of neighborhoods for r Before proceeding we 
note that for any /* £ Lj[G) there exists a pt € Co(G) such 
that 


L 


I/* - Pt I d/t < tj 4 


t The theorem nmndis 

if A X > — »- ZOO 0 open 

then the set 




|ylMx v) € O for alt* € b £ compact! 



1 1. Characters and the Dual Group 


205 


Since there must be a compact set, E k , such that p k (CE k ) = 
{0}, we have 



Assuming n such /* have been given we will denote the com- 
pact set 


n 


u Ek 

by E. 

fc— X 


Now take 


S < min , 

k = 1,2, •••,: 

x 2 H/* H* ' 

and suppose 


X € N(x o> E, 5) 

where x £ 

We now have 


]//*(*) (x(s) ~ xo(x)) dn(x) 



< / !/*(*) I I x(x) ~'Xo(x) I dp(x) 
J o 

< J i fh(x) I | x(x) — Xo(x) | dft(x) 


+ f \ft{x) | | xW — Xo(x) I dy.(x) 
Jg-E 


Hence 


< 6 i| f k ill + -J < e 


for k = 1, 2, •••, n. 


N(x o, E, S ) C TF(xo,/i, •*•,/»>«) 

which completes the proof. 



206 Element* of Abstract Harmonic Analysis 

Exercises 

1 Prove that if G is discrete r is compact, and if G is com 
pact T is discrete 

2 Let X and 1 be commutative Banach algebras with 
identities and let K be senusimple Show that if / is a 
homomorphism of X into Y, then / is continuous 

3 Let m ms and h, be Bore! measures on a locally compact 
Hausdorff space X such that = hi + « Prove that the 
regularity of any two of them implies that of the third 

4 Let G be a locally compact topological abelian group 

Show that G is the union of a disjoint collection of <r- 
compact sets |F«} and each <r-compact subset of G is 
contained in the union of a countable subcollection of the 
U?,j X is called a-compact if X = F, compact 

References 

} Naim&rk Normed Ring) 

2 Rurhn Fourier Analyst* on Groups 



CHAPTER 12 


Generalization of the 


Fourier Transform 
to LifGj and L 2 (G) 


In this chapter we will define the generalization of the 
Fourier transform; a result toward which our previous work 
has been directed. First we shall define the Fourier transform 
of any function in L t (G ) where G is a locally compact, 
abelian, topological group, ( G will carry this connotation 
throughout the rest of this chapter) and will then briefly 
sketch the Fourier-Stieltjes transform after some necessary 
notions about complex measures have been introduced. In 
addition to briefly discussing some peripheral material that 
arises, the Fourier transform over Li(G) will then be ex- 
tended to L 2 (G). 

The Fourier Transform on Ii(G) 

Having shown V to be a group and having defined a 
topology on T, we now wish to show that T is a topological 
group. Hence it must be shown that the operations of taking 
inverses and that of multiplication are continuous. 

First we will show that multiplication is continuous and 
to this end suppose N(xox'o> E, <0 is a neighborhood of xox'o- 
We must now find neighborhoods of xo and xo> Nj and N:, 
such that 

NiN 2 C N(xox' 0 , E, e) 

and we contend that N( X o, E, e/2) and N (xj, E, e/2) are 
suitable neighborhoods. To demonstrate the desired inclu- 
sion let 

X 6 N( x o, E, e/2) 

x' € W( X J, E, e/2) 

20 7 


and 



208 


Elements of Ab itroct Harmonic Analy* $ 


and consider 

txGrix'Cy) - x»(y)xi(y) t 

= lx(y)x'(y) - xo(y)x'(y) + xo(y)x'(y) - xo(y)xJ(y) 1 
- Ix'(y)(x(y) - x»(y)) + (x'(y) - xJ(y))xo(y) I 
£ I x'(y) 1 1 x(y) - x<i(y) | + 1 x'(y) - xj(y) 1 1 xa(y) I 

< i + \ ** * for a11 v e E 

This implies 

xx' € -^(xjxJ, B, *) 

and proves that multiplication is continuous 
To show that taking inverses is continuous it is first noted 
that 

x(y)x(y) - 1 for any y 6 G 

implies that 

(x(y)>-‘ - x-Hy> - x(v) 
or 

x -1 “ x 

Now given a neighborhood of xo” 1 . N()&\ B, t ), we must 
find a neighborhood of xo which is mapped into it We claim 
that N{xt>, E, *) does the trick, for suppose x £ N(xo. E, «) 
Then 

I x~'(y) - xrHy) I - I x(y) - xo(y) I 
= I x(y) - xa(y) l 
* lx(y) - xo(y) I < * 
which implies x -1 6 N(xt\ E, t) 

Having proved that r is a locally compact, abelian, top- 
ological group v.e can proceed to define the Fourier transform 
off € L\(G) as 


fix) = Jf(y)xiv) d»{y) 


(i) 



1 2. Generalization of the Fourier Transform 


209 


In view of the 1-1 correspondence between r and the maximal 
ideals, M [excluding Li(G ) ], if M £ M and "corresponds” 
to x we have 


fix) = f(M). (la) 


Further, since we threw the weak topology on r, the above 
functions are each continuous functions of their respective 
arguments.! 

Some immediate consequences of (1) or (la) are, for 
fi, h € L\(G) and X £ C, 


/ \ 


+ / 2 )(x) = fi(x) + Mx) 

(2) 

/\ 


m (x) = x/i(x) 

(3) 

/\ 


fi*Mx) — /i(x)/s(x) - 

(4) 


In Chapter 6 the notion of an algebra with involution was 
defined. In the case of L\(G) if we take 

f*(y) = /Or 1 ) 

as the involution operation, it can easily be verified as noted 
on p. 188 that it is an algebra with involution. We now have 


/*(x) = ff(y~ 1 )x(y)d l i(y). 

G being abelian, however, implies that, for any measurable 
Be t, E, n(E) = n(E ~ l ) . Replacing y by y~ x in the integrand 
of the above expression, we obtain 


/*(x) = ff(y)x(T 1 ) dn{y). 

•>a 

But, since x 6 T, 

xClT 1 ) = x -1 (f) = x(y)', 

hence 

f*(x) = 7(x) 


t See footnote to Theorem 5, Chapter 11, p. 203. 



212 


Elements of Abstract Harmonic Analyui 


It can be shonn (cf Halmos [S3) that { m I is a measure 
With this result in mind we define p to be regular if | p J is 
regular (See p 215 for the definition of regularity ) 

Wen ill denote by B( Y) all those complex, regular, Bore! 
measures, p, such that 

ImI(K) < « 

Defining addition and scalar multiplication of the set func- 
tions in B(Y) bj 

1 m Pi € B(Y), E € 5, 

(#»i + =» pi(E) + Pi{E) 

and 

2 A 6 C p£ B(Y), E e S, 

(\p)(E)=\ p(E) 

we now claim that B{ Y) is a Banach space with respect to 
the norm 

!MI = Utm p£B(Y) 

Some additional measure theoretic results about complex 
measures mil non be stated after making the following 

Definition A signed measure is a countably additive, ex- 
tended real-valued set function p on the class S (in general, 
on a ff-rmg of subsets of an abstract set) such that n(0) =0 
and such that p assumes at most one of the values or 

Remark 1 (Jordan decomposition theorem) t Let P € BO) 
Then p can be written as 

P = Px ~ Pi + t(n» — P*) 

where the p. are measures and each p, £ BO) 

We can restate the result of this theorem by asserting that 


t One al?o has such a decomposition if n « ju«t a complex, finite- 
valued measure , in which ease the corresponding measures p, 
finite-valued (not necessarily regular) 


12. Generalization of the Fourier Transform 


213 


i can be written as 

= X + tv 

where X and „ are signed measures which belong ioB{Y). 

For ix a complex measure one defines 

jjd, - jf dpi - + '//*" "" 'll*"' 

Another import! consequence o! the above theorem^ 
that even when complex measures P ”, g0 define d. 
plement the theorem and dea wi turn our 

Having defined ”'on calsian products 

attention toward comple. , we we re dealing 

just as we spoke of product measures when we w 

with measures. . „ c \ 

Remark 2. Suppose we have two "-sure 
and (F, Sr) and let m € B{X), v 6 B(Y). 
for all A 6 Sx and all B € Sr, 

(uXr)(AXB) ~»{A)v{B). 

c v c / gee Chapter 10) is then 
The extension of y. X * to S x X r{ and „ are 

the desired measure. It can be shown that, 

regular, then so is y X v. nreceding work, having 

Again, in close analogy wi measura ble sets in the 

defined a complex measure on double integrals, 

product space, we wish now + be Fubini theorems, 

iterated integrals and new versio y. If now 

Remark 3. Let / be a Borel function on X X 
„ 6 B(X),r 6 B{Y) and one of 


f [ 1/1 d\ 

Jx J Y 




jj^f\ d\y\ d\v\, f xxy \ fl 

is finite, then 

f Mux, i-IJf'-Uf** 

JXX.Y 


Element! of Abitract Harmonic Anolyih 


214 

We wish now to show that after having introduced a 
suitable multiplication that H(G) is a Banach algel ra To 
this end let n >€ B(G) and let h bo a measurable set in 0 
Wc now take 

tm " {(x v) € GX(7|*y € 

Since /■<* must be a Bore! setf ) nG X G the follow ingdefini 
tion makes sense wo define for any measurable set ]• 

(n'»)(#)-(nX»)(J«») (7) 

Tins done we now make the follow ing 

Contortion 11(G) with respect to the operations and norri 
already indicated is a Banach algebra 

There arc now a number of things that we must v enfj in 
order to prove this and first we will show that ft •»* is a 
countably additive set function 

Countable Add tiv ty ltd y r £ B{G) and let 
F - VF 

where the F are pairwise disjoint and measurable Now wc 
can write 

(„.»)(/•) - (, •»)(u I ) - (f X')(«(/ ).) 

which since n X v is a complex product measure is equal to 

fi(i*x»)(f >, - £(*•>)<* > 

ami cstal lisbes the countable additivity of a • v Next we will 
show that a • v is regular 


f To venty that Ft > » meomrat leoccp 1S1 wbm con* dwd the 
i «tM>rc of / • e 


1 2. Generalization of the Fourier Transform 


215 


Regularity. All that will be shown here is that the conditions 
for inner regularity are satisfied; the argument for outer 
regularity following in a similar fashion, f We also may as- 
sume, making use of the Jordan Decomposition Theorem, 
that ii and v are, in addition, measures. Suppose first that E 
is a Borel set and K is any compact set contained in E. We 
now have 

(/* * v) ( K ) <<jx*v) (E) 

which implies 

sup(/x » r) (if) < ( ii*u)(E ) (incompact). (8) 

KCE 

If we can now show that for any e > 0 there exists a compact 
set, H, such that H C E and 

{n *v)(H) > (m *v)(E) - e, 

this will imply that (p *v)(E) is the least upper bound of the 
set 

( (m * v) ( K ) | K C E, K compact} 

and we can change the inequality in (8) to an equality. To 
this end we note, by Remark 2 above, that n and v being 
regular implies that ( n X v) is regular. Thus for a set 
E m , where E is measurable, and e > 0, there must exist a 
compact set F C E m such that 

ill X v){F) > (|iX v)(E ( 2) ) — e. 

t It is recalled here, that if a Borel measure, />, satisfies the condition 
that for any Borel set, E, 

(a) ii(E) = sup n(F) (F compact), 

F*ZE 

then it is said to be an inner regular Borel measure. If a Borel measure 
ft satisfies the condition that, for any Borel set, E, 

(b) n(E) = inf /t{0) (0 open). 

ECO 

Then ft is said to be an outer regular Borel measure. A Borel measure 
that satisfies both (a) and (b) is said to be a regular Borel measure. 
We note that every Haar measure is regular. 



216 Dementi of Abjtroct Hcrmonic Analyut 

Now consider the continuous mapping 
p GX G — G 
(x, y) — » xy 

and let // denote the image of F under this mapping Since 
F C E<», then 

p(F) “ // C P(£«>) C E 

or 

IIC E 

and, since the mapping is continuous, // must be compact 
In addition we ha\e p -, (//) C Hm But p~'(//) contains F 
so F C //«> Using these facts we now have 

(» ")<//> - (u X *)(//«,) 

> („X»)(f) 

>(nX ►)(£<«) - « * in»p)(E) - * 
which completes the proof 

Having shown p • v to be regular we must now show that 
it has a finite norm Then we will have the fact that (p •>>) € 
B(G) To show that B(G) is a Banach algebra however, 
we must also show that || n • » |( < |( v || ■ || v || One notes 
immediately that if this inequality holds and |i ► ( B(G) 
that this will certainly imply that (n • r) has a finite norm 
for any /j, v 6 B(G) In light of this vve will pursue proving 
only the norm inequality, knowing that we will simultane- 
ously complete the proof that B{G) is closed with respect to 
the operation • 

Proof that H a • * || < H u || * |1 r || I>ct p, »• € B(G) and 
consider 


(/*•*)(£)“ f **’) 

where, as usual, we reserve the symbol of k with a subscript 
to denote the characteristic function of the set represented 



1 2. Generalization of the Fourier Transform 


217 


by the subscript. We also have, now, 

(m * v) ( E ) = (ft X v) ( Ey >) = f k BM (x, y) d(/j X r). 

J GXG 

Since ( x , y) £ E m <=* xy £ £we can rewrite the last integral 
as 


(#* X v)(E m ) = f k E (x y) d(fi X v). 

■'GXG 

Since | m | (G) and | v | (G) are finite we can apply Remark 
3 of this chapter to justify writing the last integral as an 
iterated integral to get 

fk E d(n*v ) = f f k E (xy) dy(x) dv(y) . (9) 

Since (9) holds for any characteristic function it will cer- 
tainly also hold for any linear combination of characteristic 
functions. By defining a linear combination of characteristic 
functions to be a simple function we can restate this last 
result by saying that (9) holds for any simple function. 
Thus, using (9), we have for any simple function,/, 

[ f f(zy) dii(x) dv(y) = ffd(n*y). (10) 
-"ff -Iff •’g 

We now wish to avail ourselves of the following two theorems. 

Theorem. For any complex, bounded, Borel measurable 
function, /, there exists a sequence of simple functions {g„| 
converging uniformly to / and, conversely, every uniformly 
convergent sequence of simple functions converges to a 
bounded, Borel measurable function. 

Theorem. If a sequence of functions converges uniformly 
on a set such that the total variation of the (complex) meas- 
ure of the set is finite, then the operations of limit and inte- 
gration can be interchanged when integrating over this set. 



21 B 


Element* of Abstract Harmonic Analysts 


In \ien of the«e two theorems v>c can assert the validity 
of (10) for c\ery bounded, Borcl measurable function,/, the 
interchange on the left being justified bj noting that | n \ (G) 
and | v j ((?) are finite and being justified on the right bj the 
following equal ltj 

(m*v)(C) - (m X f) ((»<«) = (aXO(CXG) = M (C)»(0), 
and recalling the definition of the integral with respect to a 
complex measure Bearing these things in mind, non consider 

IIm'HI - U'H(G) 

which, by exercise 1 of this chapter, is equal to 

sup [fd(» •») 
i/is« I J o 1 

Using (10) we now ha\e 

II * II “ sup I f f f(zy) dn(z) dt(y) I 
l/ISi I J o J a I 

< sup f f | f(xy) I d |n | (x) d \ v | (y) 
i/isi Jo Jo 

£11 *11 II *11 

which completes the proof 

Jflaxmg shown that B(G) is a Banach algebra wc wish to 
demonstrate now that it is a Banach algebra with identity 
and we claim that the identity clement is gnen b> 

|1, e 6 EC G 

to<E) = 

[0 tU 

where t is the identity element of G taxing as an exercise 
for the student the \enfication that a, fc B{G) we shad 
demonstrate here that a. does ha\c the required property 
namclj (hat for anj n £ B(G) 


^ * fit 


<U> 



1 2. Generalization of the Fourier Transform 


219 


In view of the Jordan Decomposition Theorem (Remark 1 
of this chapter) we need only prove that (11) is satisfied for 
the case when p is a measure. In light of this let p he a meas- 
ure from B(G ) and consider 


(m *He)(E) = (p X He)(E ( ») 

= j Ve((E( 2 )) x ) dfl(x) 

where (E m ) x denotes an z-section of E ( 2 >.f We now note that 


(E m )x = {y I (X, y ) 6 E m \ 
= {y | xy 6 E ) 

= {y I V € xr l E] 

= x~'E. 


Hence the last integral becomes 

f fi e (x~ l E) dn(x) 

Jg 

which, by the definition of p e is just 

f d M {x) = y{E). QED 
J E 


The Fourier-Stieltjes Transform 

Using the same notation as in the preceding discussion let 
X € T. We will now associate with every p £ B(G) a func- 
tion jionr via 


£(x) = [ x(y) dfi{y) 
Jo 


t See the discussion of product measures in Chapter 10 for the defini- 
tion of an z-section and the justification of the equality written above. 



220 


Elements of Abstract Harmonic Analysis 


and define this to be the Founer-Shctijes transform of p We 
will now examine what this means in the framework of the 
real axis 

Example 3 Let G = R, + Then, as noted earlier in this 
lecture, due to the one-to-one correspondence between the 
continuous characters of G and the elements of G we can 
write the rouner-Stidtjcs transform of p 6 B{G) as 

Hx) - (x ( J!> 

wlicre the integration is to be performed m the Lebesgue- 
Sticltjcs sense and a(<) is of bounded \anation on the 
real axis This situation occurs because in tins framework 
anj p € B(lt) is induced b> a function of bounded variation 
on the real axis 

We now wish to extend the Fourier transform on fii((7) 
to Lt(G)f but must first introduce some preliminary notions 

Positive Definite Functions 

Definition Let // be a group and let <p be a complex-valued 
function on ll <p is said to be positne definite if 

J2 X^^y.^') > 0 

for an> 

Vi yr, • • yn 6 // and Xi X, • ■ X\ € C 
Example 4 Using the same notation as before, if / 6 Li((?) 
then the function 

v> « /' •/ 

is positne definite and continuous 

f Li(G) denotes the Banach spare of square inferrable function* 
over <7 with norm of / € U(G) defined as 

where n is a fjiven Haar measure on <7 



1 2. Generalization of the Fourier Transform 


221 


The proof that ip is positive definite is straightforward and 
is left as an exercise (see exercise 3 at the end of this chapter) . 
We prove that <p is continuous. Let/, g £ L 2 ((?) . Since C c (G) 
is dense in L 2 (<?) there exist sequences, {/„} and {g„ } , of 
functions in Co((? ) such that 

ll/»-/li*-*0 and || ffn - g || 2 -» 0. 

Now we have 

I (f*g)(x) - if n *gn)(x) | < f | S{xy-')g{y) 

- fn(xy~ l )g*{y) I dy(y) 

which equals 

f I (/(sjr 1 ) -/» (xy~'))g(y) 

J G 

+ (g(v) - 9n{y))f n {xy- x ) | dy(y) < ||/„ - /||» || g || s 

+ 11 9n — g Hz II fn II 2 

using Holder’s Inequality and, since ||/„ || 2 is bounded, we 
see that 

fn *g n -»/ *g 

uniformly. 

Moreover, denoting f x (y) — /(yx -1 ) , as in the beginning 
of Chapter 11, we have 

I ( fn * {In) (x) - (fn* ffn) (z) | 

< f l/n(zr') -hizy- 1 ) II 9n(y) I Mv) 

-'<7 

< || (/„)*-. - (/„)*-> ||i max | g n (y) |. 

yzG 

But || ( ff) x-t — {fn)z - 1 ||i can be made arbitrarily small for 
xsr 1 belonging to a suitable neighborhood of e, V, by our 
first result in Chapter 11. Thus, fn * is continuous for each 
n, and, consequently / * g is continuous. Applying this result 
to the functions/, f* £ L 2 (G) yields the desired result. 



222 Elements of Abstract Harmonic Analysis 

The definition of a positiv e definite function immediately 
implies the following three statements 

(1) If e is the identity of G and v is positive definite, then 
*>(«) > 0 

(2) Let <p be positive definite and let y £ G Then 

v>iy~ l ) - v(y) 

(3) Let $ lie positive definite let y be anj element of G, 
and let e be the identity of G Then 

I v>(y) I < s?(e) 

The first statement will be proved here to illustrate how the 
proofs of these statements go the proofs of (2) and (3) 
follow in a similar fashion and ore left as an exercise for the 
reader 

Proof of (1) All we need do to prove this statement is 
take iV = 1 j/i = e and As *= J in the definition of a positive 
definite function 

Example 5 As before fcl T denote the dual group of G and 
let f» be a measure m #(T) ic n > 0 Then 

v(y) = J x(y) <Mx) 

is continuous and positive definite, namely 

£ <Wx) 

■ .-i J r • nt-i 

.x(*») J dp(x) > o 

recalling that xW) = x(*-) 1 - x(i-) Thus v is positiv e 
definite To prov e that v> is continuous w e firvt note that if 
E is a compact «et of F then the set 

K{ r« E, t) = \x j| x(*) - x(*«) I < « for all x € / } 



12. Generalization of the Fourier Transform 


223 


is an open set in G. The proof is analogous to that, given 
after Theorem 5, of showing N(x o, E, e) is open in T. 

Next we note that 

I <p(y ) \ < [ \ x(y) I dfi(x) = f cfc(x) = m(t) = j|.^ |j. 

«' r -'r 

Now, since p is regular, there exists a compact set E C r 
such that 

n(CE) = n(T) - p(E) < 5 
where S is a preascribed positive number. Thus, 

I v(x) — \ < I \ x(x) ~ x(zo) | dn(x) 

Jo 

- / ! x(z) - x(zo) | dp(x) 

J E 


+ f ix(z) — x(xo) I dp(x) 

•>CE 

< e I! M !| + 26 

for all x £ N (x 0 , E, t) which proves the continuity of <p. 

The following theorem gives us a complete characteriza- 
tion of continuous positive definite functions in the general 
case. 

Theorem ( Bochner ). If <p is a continuous function on G, then 
<p is positive definite if and only if there exists a p £ B(T), 
P > 0, such that for all y € G 




224 Element! of Abjtroct Hormonlt Analyvi 

is positive definite and continuous The remaining half 
the proof vv ill be giv en in the appendix to this chapter 
Thus in the ca=c of the addtttv e group of real numb 
every positive definite continuous function on It can be i 
tamed by the following liebcsgue-Sticltjes integral 

v(y) - f c "daft) 

where a(() is a monotone increasing function of bourn 
variation on the real axis and com er«ol> 

In the case of the real axis however Bochners thcor 
can actually be made stronger insofar 03 we no longer requ 
that <p be continuous but only measurable i e we have i 
following result On the real axis if v is positive definite a 
measurable then there exists a monotone increasing fund: 
of bounded variation on It a(t) such that 

?(«/) *= J « *da(l) ac 

Denoting the class of all positive definite functions on 
by P and the class of all linear combinations of members 
the «et Li(0) n P by £Lj{G) n P] we can now state t 
following theorem 

Theorem A [/^(G) n P~\ is dense in L\(G) and L\(G) 

Proof t\e first observe that if / g € C e (G) then f *g 
Co(G) In particular we show that if A is the compact si 
port of / and B the compact support of g then the supp< 
of / • g is contained in AB For since / » 0 on CA and g *■ 
on CB 

f(ry~ l )e(y) - o 

except if y € B and ry~ f A or z (. Ay which in plies t! 
x € AB Consequently /• g v am hes on the complement 
the compact «ct AB T1 erefore / • p € C#(G) bv virtue of t 
proof for Example 4 



1 2. Generalization of the Fourier Transform 


225 


It follows now from Theorem 1 of Chapter 11 that the set 
of all f* * g, where f,g £ C 0 (G ) , is dense in C 0 (<?) , and, there- 
fore, dense in Li(G) since C 0 (G) is dense in Li(G). However, 

/* *g = i(/ + g)* * (/ + g) - l(f - g)* * (/ - g) 

+ ~(f 4- fff)* * (/ + ig) - \U ~ w)* * (/ - ig). 

But each of the functions 

(/± g)* * (/± g), {f± ig)* * (f ± ig) 

is positive definite and continuous. Therefore, 

f* *g e [TrfG) n p] 

and consequently, C^i(^) n ^3 ' s dense in L\{G) . 

The fact that [Li((?) n P] is dense in L 2 (G) is proved in a 
similar fashion using the fact that G^(G) is dense in L«(G) . 

Now, just as in the first two chapters, having defined the 
Fourier transform over Li(G) we now wish to focus our 
attention on the problem of recovering the given function 
from its transform. 

First let us consider the family A of all functions of the 
form/, where / 6 Li((?). We recall from our initial discus- 
sion that / is a continuous function on T. Moreover, Eqs. 
(2)-(5) of this chapter show that A is an algebra of func- 
tions, ivith respect to the customary operations, which is 
such that it contains the conjugate of any of its functions. 
In addition, since /(x) = f(M) , where M is the associated 
maximal ideal with x, the final theorem of Chapter 3 shows 
that the algebra separates points, i.e., if xi X; there exists 
an / 6 A such that/(xi) ^ /(xz)- Next, given x with associ- 
ated M, M 7^ Li(G ) , there exists an / 6 Li(G) such that 
/f I; whence, /(M) ^ 0, by the previously quoted theorem. 
In other words, for each x € r there exists an / £ A such 
that/(x) 5 ^ 0. Finally, since the one-point compactification 



226 


Element! of Abstract Harmonic Anatym 


of T amounts to adjoining M 9 *= L\{G) to M — j V,} and 
•since for / € Li(G), f(M t ) = 0 we «ee that each / € A 
vanishes at infinity Hence on the basis of the extended 
Stone-A\ cierstrass theorem (see appendix to this chapter), A 
is dense m the Banach space of all complex \a!ued continu- 
ous functions on r which \amsh at infinity, where the norm 
on the space is the usual supremum norm (\\ e recall in gen 
era! that a real or complex valued function defined on a 
locally compact IlausdorfT space is asid to rantsA of infinity 
if for c\ cry « > 0 there exists a compact set E C. \ such that 
|/(x) l < t for all * £ CE) 

\\ e apply these results to establish the follow mg 

Theorem B Let »• € B(T) If 

Jx(v) Mx) - 0 

for all y € G then ► = 0 
Proof Lct/€ Li(G) Then 

Jf(x) Mx) = J'Jfiv)x(y) My) Mx) 

- Jf(y) dM(y)fx(y) Mx) - 0 

Thus 

jt(x) Mx) - 0 

for any / € L\{G) Since by our previous discus, ion the 
family A of all / is dense in the space of all continuous func- 
tions on r which i amsh at infinity we clearly ha\ e that 

Js(x) Mx) - 0 



1 2. Generalization of the Fourier Transform 


227 


where g is any continuous function on r which vanishes at 
infinity, but the mapping 

T(g) = f g(x) d\v\( x ) 

Jr 

is a bounded linear functional on the Banach space of con- 
tinuous functions on r which vanish at infinity, since 

I T(g) | < f | g(x) I d | v\ (x) 

J r 

< ii mu ii 

where || g ||, as noted, is sup* £ r | ff(x) I- Now since T is 
bounded and T(g) =0 for all g in the Banach space, it 
follows from the Riesz representation theorem (see Appendix 
to this chapter) that v = 0. 

With this result in mind we can now proceed to the inver- 
sion formula. 

Theorem (inversion formula). Let / £ £Li (G) n P~\. Then 
fix) € Lj(r) and, if the Haar measure p of G is fixed, the 
Haar measure of T, p, can be normalized so that 

f(y) = J f(x)x(y) dp(x)- 

Proof. We note, first of all, on the basis of Bochner’s 
theorem that / can be written in the form 

f(y) = f x(y) Mx) 

J r 

where v £ B(T). Now let g £ Li(G). Then 

(ff*/)(e) = j g(y~ l )f(y) dp(y). 

J G 


( 12 ) 



226 


of Abvtrort Kamenk Anonym 


But 

ft (x) *Cx) - j f fj Cy)x(y) <fe(y) tfr(*) 

* fs( sr‘) <fe(y) / x(y) d>( x ) 

“ fv(v~ , )f( y) <fo(y). 

Consequently 

(»•/)<») -/«x)*(x) (13) 

Let A € [£a((7) n P] and apply this re«ult to the function 
y • A € Li(G) Then 

r 

((*•*)•/)(<> - ( ff*A)(x)*(x) 

J T 

“ J fffx)J(x) *(x) 

by Eq (4) of this chapter On the other hand, 

(fp-A) */) (e) - ((p-/)-A)(e) 

and since h € [Li((7) 0 ^3* another application of (13) 
implies that 

«s •/)•*> W- fe l*)/(x) ■&(*) 

where X € B{I*) is «uch that 

Mv) - Jails' ■»(*> (») 


1 2. Generalization of the Fourier Transform 


229 


Comparing these results, we see that 

J g(x)Hx) dv(x) = J g(x)fix) dx(x) (15) 

for all g where g € L\(G). Since, as observed earlier this 
family is dense in the Banach space of all continuous func- 
tions on T vanishing at infinity, ( 15) holds for all such func- 
tions, and an application of the Riesz theorem shows that 

Hx) dv{ x ) = fix ) d^(x)- (16) 

Next, let w € C 0 (r) and let E be the compact support of w. 
By our observations preceding Theorem B, we know that 
there exists a«( L\ (T) such that v(xo) ^ 0, where xo € E. 
However, Co(r) is dense in Li(r), and, therefore, for arbi- 
trary e > 0, there exists a u € C 0 (r) suc h that 

II f - «l|i < «• 

But- recalling the last theorem of Chapter 3, we have 
I Hxo) - «(xo) | < || v — u ||i < e. 

It is, consequently, clear that there exists z 0 6 Co(r) such 
that Zo(xo) t 6 0. Furthermore, by the result following Eq. 
(5) of this chapter 

/\ 

z* * Zo(xo) = I Zo(xo) I 2 > 0. 

/\ /\ 

Since z* * z 0 is continuous, z* * z 0 is positive in some neigh- 
borhood, xoUo, of xo, where t/ 0 is a suitable neighborhood 
of the identity. Now, to each x« 6 E we associate such a z a 6 
C 0 (r) and such a neighborhood U a of the identity. Clearly 

EC U XaU a 

a 

and since E is compact E C U It is then clear that, if 
h = z* * Zi + • • • + z* » z n , 

then h > 0 on E. Moreover, recalling the proof of Theorem 
A, we have that h 6 C'o(r), and by Example 4 ,h £ P, and, 
therefore to ELi((7) n P]. 



230 


Elementj of Abifroet Harmonic Anatyjii 


He now define the mapping F C*(T) — * C b)f 

F(tc) = ( vh~ l d\(x) 

J r 

where X is characterized \ia Eq (14) We first note that F 
13 well defined for if / 6 [Xi(G) n P} and if / > 0 on F, 

Jwfi' l dX(x) *= j wfi-'j-'J d\(x) 

- fvt'J-'l, *(x> 

= ( tr/ -1 «Mx) 

J T 

using Eq (1G) C!carl> F is a linear functional on C%{ T) 
Now, since h is a continuous positive definite function ne 
ha\c by Bochncr’s Theorem and Theorem E that the associ- 
ated X of Eq (14) must be a measure, whence, if tr > 0 
F{ic) > 0 

Next it is clear that one con find a tr € C»( F) and a 
suitable / £ £Li(G) n P~] «a> / £ Li(G) D P fo that the 
associated » of Eq (12) is a measure, and such that 

/ lr(x) <Mx> X s 0 

Hence tr/ £ C«(r), and 

F(trJ) = fxr/h-'d\(x) 

** /it <M x) J* 0 
J r 

m F 0 


t In the discussion below k~ l and/* 1 devoir 1/k and J// 



1 2. Generalization of the Fourier Transform 


231 


Finally, let w £ C 0 (T) and let xo € r. Let h be as above, 
i.e. h > 0 on E, the compact support of w. If we let f(y) = 
xo (y)h(y), then 

fix) = / h(y)xo{y)x(y) d/i(y) 

J o 


while 


h(xox) = I Hy) (xox) (y) dn(y) 

J G 


= f h(y)xa(.y)x(y) dn(y) 


so f(x) = h(x ox) - Therefore, by Eq. (16) 

Hx) dvix) = hix ox) dX(x) 
or 

Hx) dv(x) = h( x ) dXixo'x) I 

whence r(E) = X(xJ‘ 1 E). Thus setting v(x) = u>ixo l x) we 
have 

F(v) = fw( x ^x)ihix))- l Mx) 

•'r 


= [ w(x)ih{xox)) 1 d\(xo l x) 
J r 

= [w(x)ifix))- 1 dHxo 1 x) 

= [ wix)if(x))' 1 dvix) 

J v 





232 


Dementi of AbjfroCf Harmonic Analyst* 


Recalling the discussion of Haar measure in Chapters 8 and 
9 in particular A Sltasure Theoretic approach in Chapter 
9 ivc have that 

F(») * / te(x) dp(x) 

w here p is a Haar measure on r 
Let / be any function now in [_Li(G) n PJ and Jet u> € 
Co(T) with h as above then if * is related to / by (12) 

fw rfr(x) “ ftefh 1 rfX(x) 

J r J r 

“ F( tc/) 

“ / «/dp(x) 

J r 

Since this is true for any tc 6 Co(T) wc can now conclude 
that 

dr ^ f dp 

Butr € B{T) so 

|/ r /Jr| S^<M-IMI<« 

Hence ) € £j(r) Finally 

/(») - £t(v) <Mx) 

“ J}{x)x(y) dp(x) QED 

As an immediate consequence of tins theorem applj mg 
it to the case of the real axis viewed as an additive group 
we have 

m - />- ■<" 

Comparing this to the inversion formula obtained in 



1 2. Generalization of the Fourier Transform 


233 


Chapter 1 we immediately note that the factor 1/2? r does 
not appear here because we have normalized the Haar 
measure (Lebesgue measure in this case ) ; i.e. we have re- 
placed fi by (h/2tt) where y represents just Lebesgue measure. 

At the end of Chapter 11, we showed that the collection 
of all sets of the form N (xo, E, e), where xo <E r, e > 0 and 
E is a compact set in G constitute a fundamental system of 
neighborhoods for F. In the course of establishing the con- 
tention made in Example 5 of this chapter, we proved that 
the analogous sets N (x 0 , E, e) , where x 0 £ G, e > 0, and E is 
now a compact set of T are open. With the aid of the inver- 
sion theorem, we can now establish that these sets form a 
fundamental system for G. Namely, let V be an arbitrary 
neighborhood of e. There exists a neighborhood U of e 
such that UU* 1 C V. Moreover, by exercise 4 of Chapter 6, 
there exists a neighborhood W of e such that W C U, and, 
since G is locally compact, we may assume that F = W 
is compact. Then FF~' = WW~> C UU~' C V. If we 
denote by k? the characteristic function of the set F, and 
let / = fcp/ju(F) I/2 , then f* */, using Example 4, is con- 
tinuous and positive definite. Furthermore, 

(/**/) (x) = [ dn{y) 

J G 

= f dii(y) ( 17 ) 

J G 

= “r f k F (yx- l )kp(y) dp.(y). 

m(F) J g 

Now, k F ( y) = 1 for y £ F, and k F (yx~ l ) = 1 for yx~ l 6 F 
or x € F~ l y. Consequently h = /* */ vanishes on the comple- 
ment of FF~ l , a compact set. Therefore, h € C 0 (G) fl P C 
Li(G ) 0 P, and, by the inversion theorem, we have that 
h € Li(T), and 


fHx)x(y) d P ( X ) = Hy). 
J r 



234 


Element! of Abstract Hormone Analysis 


In particular, 

jHx) dp(x) - h(e) 

- (/* */)(0 

* —rz: [ b(y) d?(y) 

pin J c 

« 1 

by Eqx (17) We also observe that 

, /\ , 

h =/*•/= 1 / 1' >0 

Now it is a simple matter to show (<ee exercise 7 of this 
chapter) that the set function * defined bj 

r(S) - J (h(x) Mx) 

l or S a Bore] subset of T is a regular measure, whence, there 
easts a compact sub-et E of T «uch that r(CE) < t where 
we take * to be a given positive number less than one Thus 

I *= w(E) + p(CE ) < r(H) + « 
so 

/ Hx) dp(x) > 1 — « 

J E 

However, 

Hy) = f hxdp + f hx dp, 

J s J cx 

and if y € .V (e, E, *) 

|x(y) - 1 1 <« 

for x € E, so 


or 


I — Rex(y) < 
Rex(p) > 



1 2. Generalization of the Fourier Transform 


235 


Thus 

I h(y) | > \ f hx dp — ( hdp 

\ j e I J CE 

> (1 - e) f hdp - v(CE ) 

E 

> (1 - e) 2 ~ e- 

If we take, for example, e = then 
! h(y) 1 > 0 

so y 6 FF~ l C V, and, consequently, N(e, E, J) C V. Thus 
the open sets N(e, E, t) form a basis at e; hence the open sets 

XoN (e, E,e) = N (x 0 , E, e) 
form a basis for G. 

Next, we observe the following important fact: if x 0 € G 
and Xo 5 ^ e, then there exists a x 6 T such that x(xo) 5^ 1. 
Namely, since x 0 ^ e, we can find a neighborhood V of e 
such that xo £ V. Hence, using the above notation, there 
exists an N ( e , E, «) C V, so x 0 £ N ( e , U, «) , i.e. 

| x(zo) — 1 | > e for some x £ E; 

whence x(zo) 1, and the contention has been established. 

An equivalent statement to the preceding is clearly the 
following: if x lt x 2 € G and x 2 ^ xo, then there exists a x € r 
such that xfci) ^ x(* 2 ) ■ 

The next theorem will essentially provide the way by 
which we can extend the Fourier transform to L 2 (G). 

The Fourier Transform on L 2 (G) 

Theorem (Plancherel) . The mapping 

Li«?) n L 2 (G) -*■ Lz(r) 

/->/ 

is an isometry onto a dense subspace of L 2 (r) and, hence, 
can be extended to an isometry of L 2 (G ) onto L 2 (T). 



234 


Element* of Abstract Harmonic Anofyiii 


Proof We shall first show that the mapping h an isometry 
by showing that, if / € U. n U ||/iii - \\j\\ t or, to para- 
phrase this that the mapping preserves norm?) To this end 
let/ € Iq n L, and consider g = /* •/ By \irtiic of Example 
4 of this chapter it follows that g € P, hence 

g € Lin P 
But 

ll/ll! - J - //(•©« 

- //(*)/*(*-') A*(*) - s(') (18) 

where t is the identity of G By the inversion formula of this 
chapter however, we can also write g(e) as 

g (e) - fo(x)x(*) rfp(x) 

* f fir(x) dp(x) (10) 

J T 

But 

/\ 

Hx) - (/• */)(x) - I/M I’ 

so that, substituting this m (19) wc obtain 

ff(<) - ( I /(x)Na(x) - 11/Hi 

J r 

Combining this with (18) then completes the proof that 
the mapping is an isometry 

Assuming the v ahdity of the rest of the theorem wc can 
now extend the Touricr transform to Lt(G) as follows Since 
Li(G) fl Lt(G) is dense in Ly(G), for any / € L,(<7) there 
exists a sequence of functions | /.| , from ln{Q) fl ln{G), 
such that 


n in 



1 2. Generalization of the Fourier Transform 


237 


{ /„} is, therefore, a Cauchy sequence. By the isometry just 
demonstrated, the sequence { /„ J must be a Cauchy sequence 
also. Since L 2 (T) is a Banach space we can say that 

fn — *•/ 6 L 2 (T) 

and it is / that we shall define to be the Fourier transform of 
f £ Li. Having made this definition we must show that it is 
a sensible one; i.e. that/ is well defined. To this end suppose 

ii in 

/»— >f 

and H H, 

h n — * f 

where /„, /i„ € In fl L 2 for all n. Since { /„) and { A„ } each 
converge to / in the 2-norm, then 

11 fn — K U 2 — > 0 

or 

h n — >/. 

The “ontoness” of the extended mapping is seen in a 
similar fashion. 

Thus it just remains to show that the given mapping, 
/— »/, where / 6 Li((?) n L 2 (G) is a mapping onto a dense 
subspace of L 2 ( F) . We denote by A the image space of this 
linear transformation. Suppose / 6 A; consider for any 
x £ G the function 

g(x ) = x(x)f(x)- 

Define h{y) = f x (y) — f{yx~ l ). Then clearly h (E Li(G ) fl 
Li(G) and . 

ft(x) = f(y*~*)x(y) dy(y) 

J G 

= J f(y)x(yx)-dn(y) 

= xO) Jf(y)x(y) dp-iv) 

= x(*)/(x) = ff(x). 



Bemenh of Abttraet Harmonic Analytii 


23fl 

i e g «* h Therefore, A is clored under multiplication by 
xOO Now, let V be anj function in Ia(T) and let tr ~ / be 
an} function in A Suppose 

(p, tc) = / ptP dp ■» 0 
J T 

where {<p, w) denotes the customarj inner product in the 
Hilbert space Z/»(r) It follows b> what we just proied that 

/*(x)®(x)x(*) dp(x) “ 0 (20) 

for any x € O By Holder’s inequality ynfc 6 Li( F) , hence 
the set function, defined for alt Borcl sets, E, of r by 

r(E) - / v(x)Mx)x(*) Mx) 

belongs to /i(T) (*ec exercise 7 at the end of this chapter) 
But, for arbitrary y € G 

fx(v)Mx) - /n>(x)ti(x)x(xy) Mx) -0 
b} (20) , whence using Theorem B we have that r ■* 0, or 
/ *(x)tf(x)x(;r) dp(x) - 0 

1 B 

for every Borel set h of T which clearl} implies that ■* 
0 ae 

Moreover, if we set y(x) “ /(xx«) where/ <E A and x* € r, 
then the function h(y) - /(y)x«( v) € li(G) n E*(G) and 

J(x) ** ^/(y)x»(y)x(y) d/r(y) 

“ /(xx») - 0(x) 

Thus, g € A Hence suppose/ € Li(G) n /’C A(G) 0 Li{G) 



1 2. Generalization of the Fourier Transform 


239 


and / 0, then/ 6 A, and / ^ 0 by the inversion theorem, 

say /(xo) 5^ 0. If we let xt be an arbitrary element of r, then 
defining w by 

w(x) = fixxT'x o) 

we have by the above that w £ A and w(xi) = f(x o) ^ 0, 
Consequently, since the elements of A are continuous, w 
does not vanish in some neighborhood. However <pw = 0 
a.e. for all w £ A. Thus, we must have clearly that <p = 0 
a.e. Recalling that in Li(<?) we identify functions which are 
equal almost everywhere this implies that <p — 0; whence 0 
is the only element of L 2 (r) orthogonal to all of A. 

We now employ the following result from Hilbert space 
theory (cf. the text by Halmos [[4]) : If M is a closed sub- 
space of a Hilbert space X, then M = where in general 
the orthogonal complement of M, is defined as 

[y 6 X | (x, y) =0 for all x 6 M). 

It follows directly from this that if M is just a subspace of 
X, then M = M x± . In our case, with X = L 2 {T) and A = 
M, we have 

A = = (Or = L 2 (T) 

which completes the proof of the theorem. 

On the basis of the Plancherel theorem, we can show that 
the algebra, A, considered prior to Theorem B, of all func- 
tions /, / € Li(G) is just the set of all vi * <P 2 where <pi, <pt 6 
L 2 (r). For, by Parseval's formula (see exercise 10 of this 
chapter) with g in place of g, we have that, for /, g Cz L 2 (G) 


J f(y)g(y) dy(y) = Jf(x)g(x ') <Wx)- ( 21 ) 

Next, substitute g(y)xa(y) f° r {/(!/) bi (21). This yields 

j f(y)g(y)xo(y) dg(y) = jf(x)g(x ox -1 ) <Wx) (22) 
= (f*g)(x o) 



240 Hementi of Abitroct Harmonic Analyst! 

/\ 

where ire haie used the fact that ffx*(x~ l ) " f(x*x~*) ^ctr 
since every element h 6 Lj(G) is of the form fg where/, 
9 £ L*(G), we have b> Eq ( 22 ) that h(* # ) “ (/*p)(x«) 
for arbitrary x« € I\ or h « / * g but / g £ Lj(r) by Tlan- 
cherel s theorem Conversely, t! f g € Lj(r), then ( 22 ) 

show s that / * g belongs to the algebra A since f • g - fg, 
and fg € Li(G) 

We can now establish the follow mg 
Theorem C If B i* 0 is an open set in r, then there exists 
an / £ A [where /t is the algebra of all / / 6 Li(G)3 such 
that / * 0 on CD, and / ^ 0 

Proof Let E be a compact subset of R such that w ith p as 
in the inversion theorem p(E) >0 E exists by the regu* 
lanty of the Haar measure p and the fact that fi is positive 
on any nonempty open set By property 12 of Chapter C 
there exists a neighborhood 1 of e which we may assume 
compact since T is locally compact such that \ E C R 
Let k r and A* be the characteristic functions of l and E, 
respectively A y kg € Lj(r) since I and E arc compact 
Moreover k r »kg « / € A by the preceding discussion and 

(Av*A,)(x # ) «= j kvixoX ’)kt(x) <*p(x) 

But Arfxrt"') “ 0 if x«X ‘ C 1 and Mx> - 0 if x $ f 
/ - 0 on C(1 E) Z> CR Finally 

jhx> Mx) - f f M*) dp(x) 

- /l.W)*«)/wrf ’>■«*) 

- Mf>(l ) > 0, 

whence, / p* 0 

Be now wish to mate a statement about the dual group 
of T Let y be a fixed element of G and define, for x £ I\ 

f,(x) - x(y) 



12. Generalization of the Fourier Transform 


241 


We first note, by Theorem 5 of Chapter 11, that f y ( x ) is 
continuous on r. Next we note 

!/v(x) I = 1 for all x, (23) 

and, for xi, xs 6 V 

Mxixz) = (xixsM y) = xi(y)xi(y) 

= /v(xi)/u(x 2 )- (24) 

In view of this we can say that /„ is a character on r. Now, 
denoting the dual group of r by A, we can say that for each 
y € G there is some f y 6 A. Thus we have the following 
mapping: 

G —> A 

V-*fv 

In this case we also note that 

ym ~>fvJ V s 

because 

( ftnvi) (x) = x(yi Vi) = xiyOxM 

~ fvi(x)fvt(x) 

= ( fvifvz) (x) • 

Hence the mapping (24) is a homomorphism. But we can 
actually make a stronger statement about this mapping 
and will do so in the following duality theorem due to 
Pontrjagin. 

Theorem ( Pontrjagin) . The mapping 

(7 — > A 
y ~*fv, 

given above, is an isomorphism and a homeomorphism. 

Proof. We first observe that the mapping is one-to-one, 
for if y h y 2 £ G and yi ^ y 2 , then, by the discussion following 
the inversion theorem, we know that there exists a x 6 P 
such that x(Vi) X ( 2 / 2)1 i- e - ^ /«• Thus the mapping 
mentioned in the statement of the theorem which we will 
denote by F is an isomorphism of G into A. 



242 


Elements of Abjtrort Harmonic Analyili 


Next, we show that F is a homoomorphism of G onto 
F{G) Let E be a compact «et in r and consider the sets of 
the form 

V — A r (i* E , «) where i* € G, « > 0, 

and 

V — A r (tfo, E, e) where 6 A 

The sets U constitute a fundamental system of neighbor- 
hoods for G bj the discussion immediately following the 
in\ ersion theorem w htle the sets V form a fundamental sy stein 
about x» by the discussion at the end of Chapter 11, and if 
V =« N(F(x o), E «), then 

f(i/)-rnF(G), (25) 

namely if x € V then | x(*) - x(*a) l < « for all x € F, 
or l/.(x) — /« ,(x) | <« for all x € E winch means that 
F(x) -/,(! ft F(G ) Conversely, if/, € F(C) n V, then 
|/,(x) - /n(x) I < * for all X € A, which implies that 
x € V and /, =* F(x)tF(l/) Now (25) clearly implies, 
since «e aircad} Inon that F is one-to-one that F i« a 
homoomorphism of G onto F(G) 

To fini«h the proof therefore, we must ju«t show that 
F(G) ■= A We first show that F(G) is a closed subset of A 
\\c already know by the preceding that F(G) is a locally 
compact subspace of A If we let F(G) * and A* be the one- 
point compactifications of F(G) and A, rcspectit cly (see 
Chapter 7 where the one-point compactification is discussed), 
F(G)* may be considered as a subspace of A* since any 
compact set in F(G) is also compact m A But F(G)* is 
closed in A* by Theorem 4 of Chapter 5 Thus, F(G)* fl \ “ 
F(G) is closed in A 

If w c can now show that F(G) is dense in A then F(G) »■ 
b (G) ■» A, and ire will ha t e com pleted the proof If F{G) 
is not dense then F((7) “ b ((») 9* A and CF(G) r* w 
open Hence, by Theorem C, there exists an K £ A, where 
here, os u«ual /I designates those } such that / £ Ai(F), 
such that h H 0 and h *• 0 on F(G) Now 



1 2. Generalization of the Fourier Transform 243 

where h £ Li(r), and <p £ A. However, h(F(x )) = 0 for 
any x £ G. Thus, 

J h(x)x(z) dp(x ) = J h(x)fz(x) dp(x ) = 0 (27) 

for any x £ G since 

0 = k(F(x)) = fh(x)Mx) dp( x ). 

J r 

Now applying Theorem B to Eq. (27), similarly to the last 
part of the proof of the Planeherel theorem, here taking 
v{E) = / fik(x) dp(x) for an arbitrary Borel set E of T, we can 
conclude that h — 0 a.e. Therefore, h = 0, which is a con- 
tradiction, and the proof has been completed. 

We shall just draw one consequence from this duality 
theorem; namely, it follows immediately that the dual 
theorem to Theorem B is true: if p £ B(G) and if p(x) = 
Jox(y) dp(y) = 0 for all x € T, then p = 0. 

The duality theorem has a great number of further useful 
consequences and we refer the interested reader to the 
references for some of them. 

Exercises 

1. Prove that j| p || = | p | (G) = supi/|< 1 1 fof dp | where 
/ is, of course, Borel measurable. 

2. If <p is a positive defin ite function. on G, prove that, for 
x £ G, <p(x~ l ) = <p{x), and J p(x) | < <p(e). 

3. If / 6 L 2 ( G) , show that tp — f* */ is positive definite. 

4. Let f be a positive definite function on the locally com- 
pact group G. Show that if <p is continuous at the identity 
element, e, then <p is uniformly continuous on G. 

5. If H is a closed subgroup of G then prove that the dual 
group of G/H is isomorphic and homeomorphic to the 
subgroup of T consisting of those characters which are 
constant on H and its cosets. 

6. If p, v, X £ B(G), prove that 

(p * v) * X = p * (v * X) and 


p * y = v * p. 



244 Element* ef Abitrocl Harmonic Analyil* 

7 Let X 1)0 a locally compact H&usdorft space and n be a 
regular Borel measure on A' Define the sot function r 
as follows 


,(£) - f f(x) dn(x) 

where/ € Li(X,m) (see Appendix) Prove that v £ B(X) 

8 Let X be a locally compact Hausdorff space If P(A') 
designates the set of all continuous functions on A' which 
vanish at infinttj , then show that C 0 (A') is dense in 
P(A’) with respect to the supremum norm 

9 Using the same notation as m exercise 8, show that 
/ € 1 (X) if and only if tho continuous extension of /, 
which we again denote b> /, to A’*, the one point com- 
pactification of A is such that f(x*) « 0, w here x* is 
the adjoined point (see Chapter 7) 

10 Ut / g € Lt ((?) Prove Parscval’8 formula 

Jf(y)0iV) <Ml/) ” jf(x)o(x) dp(x) 


Appendix to Chapter 12 

W e list here a number of results and theorems relevant to 
Chapter 12 and also present a proof of Bochner’s theorem 

Ut A be a locally compact Ilausdorff space W c designate 
bj V(A') the Banach algebra of all complex-valued, con- 
tinuous functions on A' which vanish at infinity and with 
supremum norm 

Theorem {extended Stone-}] etertlrass theorem). Suppose A u 
a Eubalgebra of V(.Y), such that 

1 / 6 A implies / € A 

2 If x: Jj € A' and x x r* x\ then there exists an / € A 
such that /( ti) t* f(xi) 

3 If x» € A, there exists an / € A such that /(x«) r* 0 
Then A is dense in 1 (A ) 



1 2. Generalization of the Fourier Transform 


245 


Riesz Representation Theorem. Let / : V ( X ) — > C be a 

bounded linear functional. Then there exists a unique p £ 
B(X) such that 

f(y) = [ y(i) dy{t) (i) 

J X 

for y € V ( X ) . Moreover |j n || =- 1| / jj where H / 1| designates 
the norm of the bounded linear functional /. The converse 
of this theorem is also true and quite simple to prove; i.e., if 
M € B(X), then (1) defines a bounded linear functional on 
V(X). 

If X is a locally compact Hausdorff space and p a regular 
measure on X we let Li(X, it) designate all those Borel 
functions, /, on X such that 

!l/lli = / 1/14* 

is finite. For n regular it can be shown that Co(X) is dense 
in Li(X, it) with respect to the norm || ||i- Similarly one 
defines L„(X, m), 1 < P < °°, and has the corresponding 
statements. If v € B(X), we say that v is absolutely con- 
tinuous with respect to it if v{E) =0 for any Borel set E 
such that it(E) = 0. It is easy to see, that for / G Li(X, ju), 
the set function v, defined for each Borel set E as follows: 

v(E) = 

has the properties that it belongs to B(X) and is absolutely 
continuous with respect to p. The converse also holds: 

Radon-Nilcodym Theorem. If v € B(X) and is absolutely 
continuous with respect to the. measure n, then there exists 
an / G Li(X, p.) such that, 



for any Borel set E of X, 



246 


Bem*ntj of Abjtrocl HormorJc AnoIyj3» 


One frequcntfj writes f *= drfdn, or dr « fdn and speaks 
of the Radon-N ikodym dematnes {see, for example, 
Halmos £5])) 

Proof of bexhner's Theorem. Let be continuous and posi- 
tive definite W e maj assume that j* 0, and consequently 
that <r (e) = 1 recalling properties 1 and 3 of a positi\e 
definite function mentioned in the chapter It also follows 
that ^ is uniformly continuous on G bj exercise 4 of this 
chapter 

Next, choose / 6 C#(G) The function 

is uniformly continuous on EXE where E is the compact 
support of / Consequently , recalling the definition of the 
integral there exist disjoint Bore) sets E„ t « J, 2 •••, n 
such that l l^ x E ( «= E and such that 

2 /(*.)/(*>) *»(x.*7‘)i*( E,)n{E t ) 

i mi 

where x, (• E„ approximates 

/ f/(*)7(v)v>(*y~ l ) du(x) duty) ( 2 ) 

J a J o 

as clo«ely as desired The finite sum is certainly nonnegative 
since p is positne definite, whence (2) is nonnegatne for 
any / € C«(G), but C»(G) is dense in L\(G, ft) - L\(G) so, 
for any / £ Lj(G) , this is also true V, c now define F L\(G) 

C as follows 

f(S) - { f (*)*(*) dn(x) 

■'o 

F is clearly linear, and 

|F(/)I< [\f(v)\\*(v)\Mv) 

J Q 

< *uplf»(y) tll/lli (3) 


tt fit. 



1 2. Generalization of the Fourier Transform 


247 


since <p(e) = 1. Now, for/, g £ Li(G), define 

U,Q) = 

= f (/*?*) (x)<f>(x) dy(x) 

J G 


= J j f{xy~ 2 )g{y~ 2 ) v (x) dy(x ) dy(y) 
= / f(x)g(?r 1 )‘p(yz) <Mz) 


= J ff(x)g(y)<p(xir l ) dy(x) dix(y). 


It now follows from this last relation that ( /, g) is linear in 
the first argument and conjugate linear in the second argu- 
ment, and, as we already know 


(/./) = F(f */*) = J J f(x)f(y)<p(xy~ 1 ) dix(x) dy(y) > 0. 

G J G 


(/, Q) just misses being an inner product since (/, /) = 0 
does not necessarily imply that / = 0. But the properties 
that (/, g ) possesses are already sufficient to establish the 
Cauchy-Schwarz inequality. Thus 


I (f,g) | 2 < (/,/)(<?, ?)• 

Since <p is uniformly continuous there exists a neighborhood 
V of e such that 

I <p( xr 1 ) ~ <p(x) i < e 
and 

I <p(xi r 1 ) - 1 1 < t 

for y £ V and xy~ l £ V. Let U be a compact symmetric 
neighborhood of e such that U 2 C V. Then for x,y £ V, x y~ l £ 
U 2 C V and, of course, y € V. If we let g — ku/y(U ) , where 
kv is the characteristic function of U, then g € Li(G) and 



248 


Ekmeti ti of Abt fracf Harmonic Anofy»ft 


one has 

(/,<?)- ?(/) 

m fjl*) J W***) -•?(*)) Mv) M*)> 

and 

(«,g) — l «* ' fo(7/))« J 0 ~ l Ma<x)rfa<y) 

It follows that | (/, ff) — F(f) | and | (g, g) - I ( can he 
made arbitrarily 'mall, but 

|F(/> I* 

S !*•(/) - </,»>)* 

<1^(/) - (/,?) I* + </./)(?. tf) 

< I f</) - (/. If) |* +</,/) I (f f) - I I + (/,/), 

w hence 

|F(/)!*< (/./) - **(/•/•> (4) 

for an j / € J,,(G) 

Now let ft = / • /•, so ft* «= ft, and defino ft* •» ft*' 1 • ft*' 1 
for n *» 2, 3, We may apply (4) successively to the A*, 
n => I, 2, •••, since each of these functions u in L\{G) 
Doing this we get 

i r<n i * < m < 

< ... < (F(a>-)),- < 
where the last inequality follows from (3) But 
hm H A*" 11*"' - r,(A) 

by the results of Chapter 3, where r,(A) we recall is the 
spectral radius of A Thus, 

\nf)f<r.W - supM(.V) I 


12. Generalization of the Fourier Transform 


249 


by Theorem 3 of Chapter 4, but 

sup 1 h(M) | = sup | h( x ) | 

MeM xeT 

since h vanishes at infinity, i.e. h(Mo) = 0 for ikf 0 = Li(G ). 

/\ 

However, h(x) = / */* (x) = |/(x) I 2 by the consequence of 
(5) of Chapter 12. Consequently, 

| F(f) 1 < sup|/(x) |. (5) 

x«r 

We can now define a mapping, which we again denote by F, 
on the set A of all / such that / € Ln(G) into C, by 

F(f) = F{f). 

The mapping is well defined, for if / = g, (5) immediately 
implies that F(f) = F{g). The functional F(f) is of coiurse 
linear and is bounded in the supremum norm by (5). By 
the Hahn-Banaeh Theorem (cf. Naimark pj) F can be 
extended to a bounded linear functional on F(T) , the space 
of continuous complex-valued functions on I' with supremum 
norm, such that the norm of the extended functional is the 
same as that of F. We again denote the extended functional 
by F. An application now of the Riesz Theorem listed earlier 
in this appendix shows that there exists a v 6 B{T) such that 

F(g) = fff(x -1 )*(x) for all g€ F(r), 

* / r 

and 

imi = imi < i 

by (5). In particular 

F(f) = F(f) = J fix' 1 ) dv( x ) 


= j fix) d)i{x) jx{x) dvix). 



250 Efemonli of Abifraet Harmonic Anotyili 

Hence, 

Ft/) - (h !)*)«») - //(i) d M (x) /i« Mx) 

for any / € L%(G) eo 

*(*) - / x (i) *( x ) 

•'r 

for a c i € G, but both *»{*) and J7x(x) fMx) arc continu- 
ous on G, whence 


*(*) ** fx(x) *(x) 

for all x € G 
Finally, 

1 « v>(e) ~ / rf-(x) - r(D < IMt < t, 

/ r 

therefore »>(?) =■ ||f|| which readily implies that f is a 
measure on T and completes the proof of the theorem 

R» TEnENCW 

1 N nmsrlc Wormed Rirgt 

2 Hudin Founer Analyst on Group* 

3 Loomt* An Introduction to Abtlracl Harmonic Anahjtu 

4 H&Imoe Introduction to Hdbcrt Spate 

5 Hal mew M rann Theory 



Bibliography 


Bourbaki, N., Topologie Generate, Book III, Chapter I. HernSann, Paris, 
1951. 

Goldberg, R. R., Fourier Transforms. Cambridge Univ. Press, London 
and New York, 1961. 

Halmos, P. R., Introduction to Hilbert Space. Chelsea, New York, 1951. 

Halmos, P. R., Measure Theory. Van Nostrand, Princeton, New Jersey, 
1950. 

Hewitt, E., and Ross, K., Abstract Harmonic Analysis, Vol. 1. Academic 
Press, New York, 1963. 

Kelley, J., General Topology. Van Nostrand, Princeton, New Jersey, 
1955. 

Kilmogorov, A., and Fomin, A., Functional Analysis. Vol. 1: “Elements 
of the Theory of Functions and Functional Analysis.” Graylock Press, 
Rochester, New York, 1957. 

Loomis, L., An Introduction to Abstract Harmonic Analysis. Van 
Nostrand, Princeton, New Jersey, 1953. 

Naimark, M., Normed Rings. Nordhoff, Groningen, 1959. 

Gel'fand, I., Raikov, D., and Silov, G., Commutative Normed Rings 

(Amer. Math. Soc. Transl. No. 5), Amer. Math. Soc., 1957. 

Natanson, I., Theory of Functions of a Real Variable. Ungar, New York, 
1955. 

Pontrjagin, L., Topological Groups. Princeton Univ. Press, Princeton, 
New Jersey, 1946. 

Rudin, W., Fourier Analysis on Groups. Wiley (Interscience), New York, 
1962. 

Taylor, A. E., An Introduction to Functional Analysis. Wiley, New 
York, 1958. 

Van der Waerden, B., Modem Algebra. Ungar, New York, 1949. 

Zaanen, A., Linear Analysis. Wiley (Interscience), New York, 1953. 


251 



ate continuity of a measure, 

!45 

utely convergent trigono- 
netric series, space of, 26 
rence point, 62 
>ra 

mach, 25 

mpletely symmetric Banach, 
210 

sets, 125 
algebra, 125 

ith involution (star, sym- 
metric), 97, 188 
ytic vector-valued function, 29 
roximate identity, 9, 190 

inch algebra, 25 
inch space, 2 
is 

t a point, 73 
if open sets, 73 
ontinuous mapping, 62 
chner’s Theorem, 223, 246 
rel 

;omplex — measure, 211 
inner regular — measure; outer 
regular — measure, 215 
measure, 126 
regular — measure, 215 
sets, 125 

mnded linear functional, 29 

inonical mapping, 110 
artesian product (of topological 
spaces), 78-79, 85 
lauchy 

Integral Formula, 32 
Integral Theorem, 31 
lauchy-Hadamard Formula, 32 
lauchy-Schwarz Inequality, 17 


character, 193, 200 
closed set, 62 
closure, 62 
cluster point, 123 
compact, 81 
locally, 85 

compactification 
one-point, 118 
content, 163 
inner, 164 
regular, 164 
continuity 

uniform, 107 

continuous 
mapping, 62 

partition of the identity, 121 
convergence 

in a topological space, 123 
of a generalized sequence, 116 
convolution, 6, 180 
coset, 50 
covering, 81 
finite, 81 
open, 81 
countability 

first axiom of, 73 
second axiom of, 74 
countably additive class, 125 

Daniell extension, 158, 166 
directed set, 116 
discrete topology, 61 
dominated convergence theorem, 
16 

double integral, 178 
dual group, 193 
duality theorem, 241 

essential supremum (ess sup), 196 


253 



254 


lode* 


nteaibd Stone-W eierstrasn Theo- 
rem, 2-14 

extension theorem fTietxe), J1R 

factor group, 114 
FatouV Lemma, )6 
finite intersection property , S3 
Fourier Transform 
on £,(-«>. »), Iff 
-> 19-20 
on ti(G), 207 ff . 210 ff 
on £»(G), 237 

Founer-Ptjpltjes Transform, 219 
ff, 224 

Fubim'a Theorem, 16, 179, 213 
fundamental system of neighbor- 
hoods 74 

Gel land 
theory, 43 (I 
topology 37 
generalised 
mfpotent element, 57 
sequence (convergence of) 116 
Cauchy sequence 116 
generated topology, 77 
group 
algebra 27 
dual, 193 
factor, 114 
general linear, 99 
homogeneous 103 
quotient, 114 
regular, 106 
topological, PS 
ununodular, 99 

Haar 

covering function, 130 
integral, 12$, 129, 172 
measure, 126, 169, 172 
Il&usdorff space, SO 
Hildw’i inequality, 17 
homeomorphism, 62 
homogeneous, 103 


ideal, 4S 
maximal, 49 
principal, W 
proper, 43 
identity 

continuous partition of, 121 
induced 
measure, 12S 
topology, 78 
interior point, 71 
inversion formulas 
L»(-«, »), 5 
»),21 
Li(G) 227 
involution 
algebra with, 97 
isomorphism, 123 
iterated integral, 178 

•Jordan Decomposition Theorem, 

212 

lattice 

linear vector, 159 
Lcbesgue point, 5 
limit point, 123 
Liouville • Theorem 30 
locally compact, 85 
lower aemieontinuous functior 
166 

maximal ideal 49 
measurable 
function, 12S 
mS IG1 
set, 168 

transformation 127-128 
measure 

a!*olute continuity of, 243 
complex Bore!, 211 
outer, 64 
product, 176 ff 
signed, 212 



Index 


255 


Minkowski inequality, 17 
modular function, 187 

natural mapping, 110 
neighborhood, 66 ff. 
fundamental system of, 74 
symmetric, 104 
topology, 66 ff. 
nilpotent element, 57 
normal space, 117 
normal subgroup, 110 
normed algebra, 25 

One-point Compactification, 118 
open sets, 61 

in terms of neighborhoods, 67 
open mapping, 111 
outer measure, 164 

Parseval’s Theorem, 20, 244 
Plancherel’s Theorem, 24, 235 
Pontrjagin Duality Theorem, 241 
positive definite function, 220 
positive linear functional, 158 
power set, 61 
principal ideal, 50 
product measures, 176 ff. 
product space, 78, 79 
projection mapping, 84 
proper ideal, 48 
pth power summable, 1 

quotient 
algebra, 50 
group, 114 

radical, 56 

Radon-Nikodym Theorem, 245 
regular_ 
content, 163 

measure (inner, outer), 215 
point, 37 

topological space, 106 
relative topology, 78 
Representation Theorem, 160 
Riesz, 245 


resolvent operator, 41 
Riemann-Lebesgue Lemma, 4 
right Haar measure, 126, 172 ff. 
ring of sets, 125 

semisimple, 57 
separation axioms, 79 ff. 
o-algebra, 125 
<r-bounded, 164 
tr-ring, 125 
simple function, 217 
spectral radius, 42 
spectrum of an element, 37 
star (symmetric) algebra, 97 
Stone-Weierstrass Theorem, 244 
stronger topology, 77 
subbasis (subbase), 75 
subgroup, 109 
closed, 110 
normal, 114 
subspace topology, 78 
summable set, 168 
support of a function, 107 
symmetric neighborhood, 104 

Tietze Extension Theorem, 118 
ToDelli-Hobson Theorem, 17 
topological divisor of zero, 53 
topological group, 98 
homogeneous, 103 
locally compact, 107 
normal subgroup, 114 
regular, 106 
subgroup of, 109 
topological space, 61 ff. 
topology, 61 
discrete, 61 
Gel’fand, 87 
generated, 77 
induced, 78 
relative (subspace), 78 
trivial, 61 
Tychonoff, 85 
weak, 78 

weaker (stronger), 77 



256 


Inden 


total variation, 211 
Tychonoff 
Theorem, 85 
topology, 85 


weak topology, 78 
weaker topology, 77 

%\iener Theorem, CO 

i -section, 177 


uniformly continuous, 107 

ummodular group, 90 
unit, 34 

Urysohn’a Lemma, 118 


y -«ection, 177 

*ero set, 167 
Zorn’s Lemma, 49